Maximum Hamming distance among a list of padded strings

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
18
down vote

favorite
2












The Hamming distance between two strings of equal length is the number of positions at which the corresponding characters are different. If the strings are not of equal length, the Hamming distance is not defined.



Challenge



Write a program or function that finds the largest Hamming distance from among all pairs of strings from a list of strings, padded as required according to the rules described below.



The characters will be from within a-zA-Z0-9.



The strings may not be equal in length, so for each comparison the shorter string has to be padded as follows:



  • wrap the string from the beginning as many times as needed to match the required length

  • change the cases of the letters each odd time wrapping (1st, 3rd, 5th, etc.)

  • leave things outside a-zA-Z unchanged when wrapping

For example, let's say you need to pad the 5 character string ab9Cd so that it ends up with 18 characters. You would end up with:



ab9CdAB9cDab9CdAB9
^^^^^ ^^^


with ^ added underneath the 1st and 3rd wraps to highlight to case changes.



Input/Output



Input/output format is flexible. You can assume the input has at least two strings, and that all strings will have at least one character.



The output is an integer.



Rules



This is code-golf. Standard rules apply.



Test cases



[ "a", "b" ] => 1
[ "a", "b", "c" ] => 1
[ "a", "a", "c" ] => 1
[ "abc", "abcd" ] => 1
[ "abc12D5", "abC34d3", "ABC14dabc23DAbC89d"] => 17
[ "a", "Aaa", "AaaA", "aAaAa", "aaaaaaaaaaaaaa", "AAaAA", "aAa" ] => 8
["AacaAc", "Aab"] => 2


Reference implementation



I tested the examples with (completely ungolfed) R code that you can try here to compare any other examples you might try out with your code.







share|improve this question


















  • 1




    change the cases of the letters each odd time wrapping – Oh boy, this requirement is going to be a pain for my solution... I like the challenge, though, so +1
    – Mr. Xcoder
    Aug 13 at 19:42











  • Suggested test case: ["AacaAc", "Aab"] => 2. A purposed golf to my Jelly answer would have failed that case, but would have passes all the other ones.
    – Mr. Xcoder
    Aug 13 at 21:57











  • @ngm Excellent challenge! +1
    – Rushabh Mehta
    Aug 14 at 15:46














up vote
18
down vote

favorite
2












The Hamming distance between two strings of equal length is the number of positions at which the corresponding characters are different. If the strings are not of equal length, the Hamming distance is not defined.



Challenge



Write a program or function that finds the largest Hamming distance from among all pairs of strings from a list of strings, padded as required according to the rules described below.



The characters will be from within a-zA-Z0-9.



The strings may not be equal in length, so for each comparison the shorter string has to be padded as follows:



  • wrap the string from the beginning as many times as needed to match the required length

  • change the cases of the letters each odd time wrapping (1st, 3rd, 5th, etc.)

  • leave things outside a-zA-Z unchanged when wrapping

For example, let's say you need to pad the 5 character string ab9Cd so that it ends up with 18 characters. You would end up with:



ab9CdAB9cDab9CdAB9
^^^^^ ^^^


with ^ added underneath the 1st and 3rd wraps to highlight to case changes.



Input/Output



Input/output format is flexible. You can assume the input has at least two strings, and that all strings will have at least one character.



The output is an integer.



Rules



This is code-golf. Standard rules apply.



Test cases



[ "a", "b" ] => 1
[ "a", "b", "c" ] => 1
[ "a", "a", "c" ] => 1
[ "abc", "abcd" ] => 1
[ "abc12D5", "abC34d3", "ABC14dabc23DAbC89d"] => 17
[ "a", "Aaa", "AaaA", "aAaAa", "aaaaaaaaaaaaaa", "AAaAA", "aAa" ] => 8
["AacaAc", "Aab"] => 2


Reference implementation



I tested the examples with (completely ungolfed) R code that you can try here to compare any other examples you might try out with your code.







share|improve this question


















  • 1




    change the cases of the letters each odd time wrapping – Oh boy, this requirement is going to be a pain for my solution... I like the challenge, though, so +1
    – Mr. Xcoder
    Aug 13 at 19:42











  • Suggested test case: ["AacaAc", "Aab"] => 2. A purposed golf to my Jelly answer would have failed that case, but would have passes all the other ones.
    – Mr. Xcoder
    Aug 13 at 21:57











  • @ngm Excellent challenge! +1
    – Rushabh Mehta
    Aug 14 at 15:46












up vote
18
down vote

favorite
2









up vote
18
down vote

favorite
2






2





The Hamming distance between two strings of equal length is the number of positions at which the corresponding characters are different. If the strings are not of equal length, the Hamming distance is not defined.



Challenge



Write a program or function that finds the largest Hamming distance from among all pairs of strings from a list of strings, padded as required according to the rules described below.



The characters will be from within a-zA-Z0-9.



The strings may not be equal in length, so for each comparison the shorter string has to be padded as follows:



  • wrap the string from the beginning as many times as needed to match the required length

  • change the cases of the letters each odd time wrapping (1st, 3rd, 5th, etc.)

  • leave things outside a-zA-Z unchanged when wrapping

For example, let's say you need to pad the 5 character string ab9Cd so that it ends up with 18 characters. You would end up with:



ab9CdAB9cDab9CdAB9
^^^^^ ^^^


with ^ added underneath the 1st and 3rd wraps to highlight to case changes.



Input/Output



Input/output format is flexible. You can assume the input has at least two strings, and that all strings will have at least one character.



The output is an integer.



Rules



This is code-golf. Standard rules apply.



Test cases



[ "a", "b" ] => 1
[ "a", "b", "c" ] => 1
[ "a", "a", "c" ] => 1
[ "abc", "abcd" ] => 1
[ "abc12D5", "abC34d3", "ABC14dabc23DAbC89d"] => 17
[ "a", "Aaa", "AaaA", "aAaAa", "aaaaaaaaaaaaaa", "AAaAA", "aAa" ] => 8
["AacaAc", "Aab"] => 2


Reference implementation



I tested the examples with (completely ungolfed) R code that you can try here to compare any other examples you might try out with your code.







share|improve this question














The Hamming distance between two strings of equal length is the number of positions at which the corresponding characters are different. If the strings are not of equal length, the Hamming distance is not defined.



Challenge



Write a program or function that finds the largest Hamming distance from among all pairs of strings from a list of strings, padded as required according to the rules described below.



The characters will be from within a-zA-Z0-9.



The strings may not be equal in length, so for each comparison the shorter string has to be padded as follows:



  • wrap the string from the beginning as many times as needed to match the required length

  • change the cases of the letters each odd time wrapping (1st, 3rd, 5th, etc.)

  • leave things outside a-zA-Z unchanged when wrapping

For example, let's say you need to pad the 5 character string ab9Cd so that it ends up with 18 characters. You would end up with:



ab9CdAB9cDab9CdAB9
^^^^^ ^^^


with ^ added underneath the 1st and 3rd wraps to highlight to case changes.



Input/Output



Input/output format is flexible. You can assume the input has at least two strings, and that all strings will have at least one character.



The output is an integer.



Rules



This is code-golf. Standard rules apply.



Test cases



[ "a", "b" ] => 1
[ "a", "b", "c" ] => 1
[ "a", "a", "c" ] => 1
[ "abc", "abcd" ] => 1
[ "abc12D5", "abC34d3", "ABC14dabc23DAbC89d"] => 17
[ "a", "Aaa", "AaaA", "aAaAa", "aaaaaaaaaaaaaa", "AAaAA", "aAa" ] => 8
["AacaAc", "Aab"] => 2


Reference implementation



I tested the examples with (completely ungolfed) R code that you can try here to compare any other examples you might try out with your code.









share|improve this question













share|improve this question




share|improve this question








edited Aug 14 at 15:09

























asked Aug 13 at 19:38









ngm

2,07920




2,07920







  • 1




    change the cases of the letters each odd time wrapping – Oh boy, this requirement is going to be a pain for my solution... I like the challenge, though, so +1
    – Mr. Xcoder
    Aug 13 at 19:42











  • Suggested test case: ["AacaAc", "Aab"] => 2. A purposed golf to my Jelly answer would have failed that case, but would have passes all the other ones.
    – Mr. Xcoder
    Aug 13 at 21:57











  • @ngm Excellent challenge! +1
    – Rushabh Mehta
    Aug 14 at 15:46












  • 1




    change the cases of the letters each odd time wrapping – Oh boy, this requirement is going to be a pain for my solution... I like the challenge, though, so +1
    – Mr. Xcoder
    Aug 13 at 19:42











  • Suggested test case: ["AacaAc", "Aab"] => 2. A purposed golf to my Jelly answer would have failed that case, but would have passes all the other ones.
    – Mr. Xcoder
    Aug 13 at 21:57











  • @ngm Excellent challenge! +1
    – Rushabh Mehta
    Aug 14 at 15:46







1




1




change the cases of the letters each odd time wrapping – Oh boy, this requirement is going to be a pain for my solution... I like the challenge, though, so +1
– Mr. Xcoder
Aug 13 at 19:42





change the cases of the letters each odd time wrapping – Oh boy, this requirement is going to be a pain for my solution... I like the challenge, though, so +1
– Mr. Xcoder
Aug 13 at 19:42













Suggested test case: ["AacaAc", "Aab"] => 2. A purposed golf to my Jelly answer would have failed that case, but would have passes all the other ones.
– Mr. Xcoder
Aug 13 at 21:57





Suggested test case: ["AacaAc", "Aab"] => 2. A purposed golf to my Jelly answer would have failed that case, but would have passes all the other ones.
– Mr. Xcoder
Aug 13 at 21:57













@ngm Excellent challenge! +1
– Rushabh Mehta
Aug 14 at 15:46




@ngm Excellent challenge! +1
– Rushabh Mehta
Aug 14 at 15:46










9 Answers
9






active

oldest

votes

















up vote
7
down vote














Jelly, 20 bytes



Not really happy with it. Should be golfable, even to ~15 bytes perhaps.



LÞŒcµṁ/sḢL$ŒsÐeFn)§Ṁ


Try it online!



or Check out a test suite!



Explanation




LÞŒcµṁ/sḢL$ŒsÐeFn)§Ṁ Full program or monadic link. N = input. | Example: ["abc12D5", "abC34d3", "ABC14dabc23DAbC89d"]
LÞ Sort N by length. | [['a', 'b', 'c', '1', '2', 'D', '5'], ['a', 'b', 'C', '3', '4', 'd', '3'], ['A', 'B', 'C', '1', '4', 'd', 'a', 'b', 'c', '2', '3', 'D', 'A', 'b', 'C', '8', '9', 'd']] (in Jelly, strings are list of characters)
Ã…Â’c Unordered pairs: [x, y] for all distinct x, y in N. | [[['a', 'b', 'c', '1', '2', 'D', '5'], ['a', 'b', 'C', '3', '4', 'd', '3']], [['a', 'b', 'c', '1', '2', 'D', '5'], ['A', 'B', 'C', '1', '4', 'd', 'a', 'b', 'c', '2', '3', 'D', 'A', 'b', 'C', '8', '9', 'd']], [['a', 'b', 'C', '3', '4', 'd', '3'], ['A', 'B', 'C', '1', '4', 'd', 'a', 'b', 'c', '2', '3', 'D', 'A', 'b', 'C', '8', '9', 'd']]]
Here, by distinct, I mean at different positions. |
µ ) Map with a monadic link. |
ṁ/ Mold x like y. That is, cycle x until it reaches length y. | [['a', 'b', 'c', '1', '2', 'D', '5'], ['a', 'b', 'c', '1', '2', 'D', '5', 'a', 'b', 'c', '1', '2', 'D', '5', 'a', 'b', 'c', '1'], ['a', 'b', 'C', '3', '4', 'd', '3', 'a', 'b', 'C', '3', '4', 'd', '3', 'a', 'b', 'C', '3']]
sḢL$ Split into chunks of x's length. | [[['a', 'b', 'c', '1', '2', 'D', '5']], [['a', 'b', 'c', '1', '2', 'D', '5'], ['a', 'b', 'c', '1', '2', 'D', '5'], ['a', 'b', 'c', '1']], [['a', 'b', 'C', '3', '4', 'd', '3'], ['a', 'b', 'C', '3', '4', 'd', '3'], ['a', 'b', 'C', '3']]]
ŒsÐe Swap the case of even-indexed chunks (1-indexed). | [[['a', 'b', 'c', '1', '2', 'D', '5']], [['a', 'b', 'c', '1', '2', 'D', '5'], ['A', 'B', 'C', '1', '2', 'd', '5'], ['a', 'b', 'c', '1']], [['a', 'b', 'C', '3', '4', 'd', '3'], ['A', 'B', 'c', '3', '4', 'D', '3'], ['a', 'b', 'C', '3']]]
F Flatten. | [['a', 'b', 'c', '1', '2', 'D', '5'], ['a', 'b', 'c', '1', '2', 'D', '5', 'A', 'B', 'C', '1', '2', 'd', '5', 'a', 'b', 'c', '1'], ['a', 'b', 'C', '3', '4', 'd', '3', 'A', 'B', 'c', '3', '4', 'D', '3', 'a', 'b', 'C', '3']]
n Vectorized inequality with y. | [[[0, 0, 1, 1, 1, 1, 1]], [[1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]], [[1, 1, 0, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1]]]
§ After ending the map, sum each bool (0 or 1) array. | [[5], [17], [14]]
Ṁ Maximum. | 17





share|improve this answer






















  • I'm totally not familiar with Jelly, but I think you can omit LÞ and still get the same maximum at the end.
    – Chas Brown
    Aug 13 at 21:41











  • @ChasBrown Ugh, no, I should need that. Otherwise, instead of padding the shortest to the length of the longest one, ṁ/ would instead trim the longest to the length of the shortest one in some cases, which is not what we want.... I guess the test cases are too well chosen (and this is a rather unfortunate coincidence)...
    – Mr. Xcoder
    Aug 13 at 21:48











  • @ChasBrown As an example, try ["AacaAc", "Aab"].
    – Mr. Xcoder
    Aug 13 at 21:57










  • Ah yes, I see... I need to learn me some more Jelly... :)
    – Chas Brown
    Aug 13 at 22:02

















up vote
5
down vote














Python 2, 86 bytes





lambda a:max(sum(x!=y for x,y in zip((s+s.swapcase())*len(t),t))for s in a for t in a)


Try it online!



Given two strings, s,t, zip((s+s.swapcase())*len(t),t)) will be a list of tuples of length len(t) since zip truncates to the shortest iterable. If len(s)<len(t), then this "pads out" s with the desired case swapping and we calculate the sum of differing characters.



If len(t)<=len(s), then the resulting sum will be less than or equal to the sum if we were evaluating t,s; so it has no effect on the resulting max in that case.






share|improve this answer






















  • You can use y!= instead of !=y to save 1 byte
    – Mr. Xcoder
    Aug 13 at 20:43










  • @Mr. Xcoder: Thx, but I ended up drastically reworking my solution...
    – Chas Brown
    Aug 13 at 21:10

















up vote
3
down vote














JavaScript (Node.js), 111 bytes





a=>a.map(m=S=>a.map(s=>B(S).map((c,k)=>m=(c^(c=B(s)[k%(l=s.length)])^(k/l&c>9)<<5&&++x)<m?m:x,x=0)),B=Buffer)|m


Try it online!






share|improve this answer



























    up vote
    3
    down vote














    Jelly, 19 bytes



    WṁŒsÐeF=ċ0
    LÞŒcç/€Ṁ


    Try it online!



    LÞŒcç/€Ṁ
    LÞ Sort by length
    Ã…Â’c unordered pairs
    € to each of the pairs
    ç/ find the hamming distance with molding and swapping case (helper link)
    Ṁ maximum

    WṁŒsÐeF=ċ0
    W wrap the shorter string
    ṁ repeat this string once for each character in the second string
    Ðe for every other repeated string
    Ã…Â’s swap case
    F flatten
    = characterwise equality check between the two strings. If the first
    string is longer, the leftover characters are appended to the result.
    e.g. 'abABab' and 'xbA' give [0,1,1,'B','a','b']
    ċ0 count the number of 0s, giving the Hamming distance.





    share|improve this answer





























      up vote
      2
      down vote














      Ruby, 89 82 bytes



      Creates the cross-product of the input list against itself before calculating the Hamming distance of each pair, using a duplication method similar to Chas Brown's answer. Ruby can't zip strings together or add booleans without additional overhead, though, so it becomes necessary to iterate through the pair of strings manually instead.



      -7 bytes from GB.





      ->aa.product(a).maps,t.max


      Try it online!






      share|improve this answer


















      • 1




        82 bytes
        – G B
        Aug 14 at 11:27

















      up vote
      2
      down vote














      Java 10, 748 740 667 666 616 bytes



      This has to be the most dense and unreadable, yet the longest golf I ever came up with.



      Call method h(String) with an explicit array (no var args): eg,



      h(new String "a", "b", "c");


      returns 1.



      char e(boolean w,char c)96<c&c<123)?c^32:c);String p(String s,int l)var p="";int m=s.length(),n=l/m,r=l%m,i=0,j=0;var w=1<0;for(;i<n;++i,w=!w)for(char c:s.toCharArray())p+=e(w,c);for(;j<r;)p+=e(w,s.charAt(j++));return p;int d(String s,String t)int l=s.length(),n=0,i=0;for(;i<l;)if(s.charAt(i)!=t.charAt(i++))++n;return n;int h(String s,String t)int l=s.length(),m=t.length();return l>m?d(s,p(t,l)):l<m?d(p(s,m),t):d(s,t);int h(Strings)int l=s.length,i=0,j;intn=new int[l*l];for(;i<l;++i)for(j=i;++j<l;)n[i*l+j]=h(s[i],s[j]);return java.util.Arrays.stream(n).max().getAsInt();


      You can Try It Online!



      Ungolfed and commented:



      // Encode the character (swap case)
      char e(boolean w, char c) 96 < c & c < 123) ? c ^ 32 : c);


      // Pad the string to desired length
      String p(String s, int l)
      var p = "";
      int m = s.length(), n = l / m, r = l % m, i = 0, j = 0;
      var w = 1 < 0;
      for (; i < n; ++i, w = !w)
      for (char c : s.toCharArray())
      p += e(w, c);
      for (; j < r;)
      p += e(w, s.charAt(j++));
      return p;


      // Calculate the actual hamming distance between two same-length strings
      int d(String s, String t)
      int l = s.length(), n = 0, i = 0;
      for (; i < l;)
      if (s.charAt(i) != t.charAt(i++))
      ++n;
      return n;

      // Pad the strings as needed and return their hamming distance
      int h(String s, String t)
      int l = s.length(), m = t.length();
      return l > m ? d(s, p(t, l)) : l < m ? d(p(s, m), t) : d(s, t);


      // Dispatch the strings and gather their hamming distances, return the max
      int h(String s)
      int l = s.length, i = 0, j;
      int n = new int[l * l];
      for (; i < l; ++i)
      for (j = i; ++j < l;)
      n[i * l + j] = h(s[i], s[j]);
      return java.util.Arrays.stream(n).max().getAsInt();



      I know a better solution can be achieved, especially for the string pairing part.



      EDIT: shave off 8 bytes by changing the size of the int array in hammingDistance() to the square of the numbe of strings given. It also fixes an ArrayIndexOutOfBounds thrown in one of the test cases.



      EDIT 2: Saved 33 bytes thanks to Kevin Cruijssen's comments: class declaration removed, names shortened to 1 char, operators changed, etc.



      EDIT 3: Save 1 byte and reach Satan-approved score by changing method with var-arg to array.



      EDIT 4: Save another 50 bytes thanks to Kevin Cruijssen, again: update Java version from 8 to 10 to use var keyword, removed StringBuilder instance, etc.






      share|improve this answer


















      • 1




        I don't have a lot of time, but some basic things to golf: Drop the class, only the methods is enough. Change all method and variable names to single bytes.. So instead of hammingDistance use d or some other unused variable. Most of your && can be & and || can be |. c^' ' can be c^32. boolean w = false; can be boolean w=0>1;. i=0 in the loop initialization can be removed and change the ,i,j to ,i=0,j. ++j can be removed and ++ can be added to the .charAt(j++). .toString() can be +"". for(j=i+1;j<l;++j) can be for(j=0;++j<l;). Etc. etc.
        – Kevin Cruijssen
        Aug 14 at 13:38






      • 1




        Tips for golfing in Java and Tips for golfing in <all languages> might be interesting to read through as well. :)
        – Kevin Cruijssen
        Aug 14 at 13:41










      • Thanks! That's some nice byte-lifting. Thanks for the links too, I'm taking a look at it and will edit asap!
        – joH1
        Aug 14 at 13:47






      • 1




        Upvoted for the Satan-approved score. xD Some more small things: StringBuilder can be StringBuffer (if you switch to Java 10 it could be var b=new StringBuffer(l);. The boolean and char can then also be var. If you don't have Java 10 locally, it is available on TIO). In addition, for(;i<n;++i)for(char c:s.toCharArray())b.append(e(w,c));w=!w; can be for(;i++<n;w=!w)for(char c:s.toCharArray())b.append(e(w,c));. And I'm pretty sure you can remove the StringBuffer completely and just use String and += instead of append.
        – Kevin Cruijssen
        Aug 14 at 15:13











      • Man, some many months of clean code and good coding practices made me forget how to even golf! I'll update my answer and include TIO.
        – joH1
        Aug 16 at 7:15

















      up vote
      1
      down vote














      05AB1E, 33 29 bytes



      Ćü)€é©εćDš«s`g∍}®€¤‚ø€ζ€€Ë_Oà


      Try it online or verify all test cases.



      Can most likely be halved in byte-count, but it works..



      Explanation:





      Ć # Enclose the input-list (adding the first item to the end of the list)
      # i.e. ['ABC1','abcD','abCd32e'] → ['ABC1','abcD','abCd32e','ABC1']
      ü) # Pair-vectorize each of them
      # i.e. ['ABC1','abcD','abCd32e','ABC1']
      # → [['ABC1','abcD'],['abcD','abCd32e'],['abCd32e','ABC1']]
      ێ # Sort each pair by length
      # i.e. [['ABC1','abcD'],['abcD','abCd32e'],['abCd32e','ABC1']]
      # → [['ABC1','abcD'],['abcD','abCd32e'],['ABC1','abCd32e']]
      © # Store this list in the register to re-use later on
      ε } # Map each inner list in this list to:
      ć # Head extracted
      # i.e. ['abcD','abCd32e'] → 'abcD' and ['abCd32e']
      DÅ¡ # Duplicate it, and swap the capitalization of the copy
      # i.e. 'abcD' → 'ABCd'
      « # Then merge it together
      # i.e. 'abcD' and 'ABCd' → 'abcDABCd'
      s` # Swap so the tail-list is at the top of the stack, and get it's single item
      # i.e. ['abCd32e'] → 'abCd32e'
      g # Get the length of that
      # i.e. 'abCd32e' → 7
      ∍ # Extend/shorten the string to that length
      # i.e. 'abcDABCd' and 7 → 'abcDABC'
      ® # Get the saved list from the register again
      €¤ # Get the tail from each
      # i.e. [['ABC1','abcD'],['abcD','abCd32e'],['abCd32e','ABC1']]
      # → ['abcD','abCd32e','abCd32e']
      ‚ # Pair it with the other list
      # i.e. ['ABC1','abcDABC','ABC1abc'] and ['abcD','abCd32e','abCd32e']
      # → [['ABC1','abcDABC','ABC1abc'],['abcD','abCd32e','abCd32e']]
      ø # Zip it, swapping rows / columns
      # i.e. [['ABC1','abcDABC','ABC1abc'],['abcD','abCd32e','abCd32e']]
      # → [['ABC1','abcD'],['abcDABC','abCd32e'],['ABC1abc','abCd32e']]
      €ζ # And then zip each pair again
      # i.e. [['ABC1','abcD'],['abcDABC','abCd32e'],['ABC1abc','abCd32e']]
      # → [['Aa','Bb','Cc','1D'],['aa','bb','cC','Dd','A3','B2','Ce'],['Aa','Bb','CC','1d','a3','b2','ce']]
      € # Then for each inner list
      € # And for each inner string
      Ë # Check if all characters are the same
      # i.e. 'aa' → 1
      # i.e. 'cC' → 0
      _ # And inverse the booleans
      # i.e. [['Aa','Bb','Cc','1D'],['aa','bb','cC','Dd','A3','B2','Ce'],['Aa','Bb','CC','1d','a3','b2','ce']]
      # → [[1,1,1,1],[0,0,1,1,1,1,1],[1,1,0,1,1,1,1]]
      O # Then sum each inner list
      # i.e. [[1,1,1,1],[0,0,1,1,1,1,1],[1,1,0,1,1,1,1]] → [4,5,6]
      à # And take the max as result
      # i.e. [4,5,6] → 6





      share|improve this answer



























        up vote
        1
        down vote













        Java 11, 387 bytes





        a->int l=a.length,i=l,t,j=0,C=new int[l];var p=new String[l][2];for(;i-->0;p[i][0]=a[t>0?i:j],p[i][1]=a[t>0?j:i])t=a[i].length()<a[j=-~i%l].length()?1:0;i=0;for(var P:p)var s="";for(var x:P[0].getBytes())s+=(char)(x>64&x<91for(int c:C)j=c>j?c:j;return j;


        Try it online. (NOTE: Since Java 11 isn't on TIO yet, String.repeat(int) has been emulated as repeat(String,int) for the same byte-count.)



        Explanation:



        a-> // Method with String-array parameter and integer return-type
        int l=a.length, // Length of the input-array
        i=l, // Index-integer, starting at the length
        t,j=0, // Temp-integers
        C=new int[l]; // Count-array the same size as the input
        var p=new String[l][2]; // String-pairs array the same size as the input
        for(;i-->0 // Loop `i` in the range [`l`, 0)
        ; // After every iteration:
        p[i][0]= // Set the first String of the pair at index `i` to:
        a[t>0?i:j],// The smallest of the `i`'th or `j`'th Strings of the input-array
        p[i][1]= // And set the second String of the pair at index `i` to:
        a[t>0?j:i])// The largest of the `i`'th or `j`'th Strings of the input-array
        t=a[i].length()< // If the length of the `i`'th item is smaller than
        a[j=-~i%l].length()?// the length of the `i+1`'th item
        // (and set `j` to this `i+1` with wrap-around to 0 for the last item
        1 // Set `t` to 1 as flag
        : // Else:
        0; // Set `t` to 0 as flag
        // We've now created the String pairs, where each pair is sorted by length
        i=0; // Reset `i` to 0
        for(var P:p) // Loop over the pairs
        var s=""; // Temp-String starting empty
        for(var x:P[0].getBytes())
        // Loop over the characters of the first String of the pair
        s+= // Append the temp-String with:
        (char)(x>64&x<91 // After every iteration of the outer loop,
        // increase `i` by 1 for the next iteration
        for(int c:C) // Now loop over the calculated counts
        j=c>j?c:j; // And set `j` to the maximum
        return j; // And finally return this maximum `j` as result





        share|improve this answer



























          up vote
          1
          down vote














          R, 173 bytes





          function(x,U=utf8ToInt,N=nchar)max(combn(x,2,function(z,v=z[order(N(z))])sum(U(substr(Reduce(paste0,rep(c(v[1],chartr('A-Za-z','a-zA-Z',v[1])),n<-N(v[2]))),1,n))!=U(v[2]))))


          Try it online!



          @ngm : I tried my best to golf your code (with my heavy customizations of course) but, as you well know, R is not very golfy in manipulating strings :P






          share|improve this answer






















          • I bet this can be sub 150 bytes, but I'm not sure how just yet.
            – Giuseppe
            Aug 15 at 17:50










          • @Giuseppe: I suspect that too... but I'm not really good in writing short strings manipulation codes and R doesn't help me very much either :D
            – digEmAll
            Aug 15 at 18:56










          • @digEmAll I'm not going to try to solve my own challenge, but few possibilities might include outer to get all the combinations, and doing modular arithmetic on the code points in lieu of chartr.
            – ngm
            Aug 16 at 0:33










          • @ngm: possible...I discarded the arithmetic approach because I couldn't find a short solution/formula to change the case for letters without touching the numbers...
            – digEmAll
            Aug 16 at 6:18










          Your Answer




          StackExchange.ifUsing("editor", function ()
          return StackExchange.using("mathjaxEditing", function ()
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["\$", "\$"]]);
          );
          );
          , "mathjax-editing");

          StackExchange.ifUsing("editor", function ()
          StackExchange.using("externalEditor", function ()
          StackExchange.using("snippets", function ()
          StackExchange.snippets.init();
          );
          );
          , "code-snippets");

          StackExchange.ready(function()
          var channelOptions =
          tags: "".split(" "),
          id: "200"
          ;
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function()
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled)
          StackExchange.using("snippets", function()
          createEditor();
          );

          else
          createEditor();

          );

          function createEditor()
          StackExchange.prepareEditor(
          heartbeatType: 'answer',
          convertImagesToLinks: false,
          noModals: false,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: null,
          bindNavPrevention: true,
          postfix: "",
          onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          );



          );













           

          draft saved


          draft discarded


















          StackExchange.ready(
          function ()
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcodegolf.stackexchange.com%2fquestions%2f170581%2fmaximum-hamming-distance-among-a-list-of-padded-strings%23new-answer', 'question_page');

          );

          Post as a guest






























          9 Answers
          9






          active

          oldest

          votes








          9 Answers
          9






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          7
          down vote














          Jelly, 20 bytes



          Not really happy with it. Should be golfable, even to ~15 bytes perhaps.



          LÞŒcµṁ/sḢL$ŒsÐeFn)§Ṁ


          Try it online!



          or Check out a test suite!



          Explanation




          LÞŒcµṁ/sḢL$ŒsÐeFn)§Ṁ Full program or monadic link. N = input. | Example: ["abc12D5", "abC34d3", "ABC14dabc23DAbC89d"]
          LÞ Sort N by length. | [['a', 'b', 'c', '1', '2', 'D', '5'], ['a', 'b', 'C', '3', '4', 'd', '3'], ['A', 'B', 'C', '1', '4', 'd', 'a', 'b', 'c', '2', '3', 'D', 'A', 'b', 'C', '8', '9', 'd']] (in Jelly, strings are list of characters)
          Ã…Â’c Unordered pairs: [x, y] for all distinct x, y in N. | [[['a', 'b', 'c', '1', '2', 'D', '5'], ['a', 'b', 'C', '3', '4', 'd', '3']], [['a', 'b', 'c', '1', '2', 'D', '5'], ['A', 'B', 'C', '1', '4', 'd', 'a', 'b', 'c', '2', '3', 'D', 'A', 'b', 'C', '8', '9', 'd']], [['a', 'b', 'C', '3', '4', 'd', '3'], ['A', 'B', 'C', '1', '4', 'd', 'a', 'b', 'c', '2', '3', 'D', 'A', 'b', 'C', '8', '9', 'd']]]
          Here, by distinct, I mean at different positions. |
          µ ) Map with a monadic link. |
          ṁ/ Mold x like y. That is, cycle x until it reaches length y. | [['a', 'b', 'c', '1', '2', 'D', '5'], ['a', 'b', 'c', '1', '2', 'D', '5', 'a', 'b', 'c', '1', '2', 'D', '5', 'a', 'b', 'c', '1'], ['a', 'b', 'C', '3', '4', 'd', '3', 'a', 'b', 'C', '3', '4', 'd', '3', 'a', 'b', 'C', '3']]
          sḢL$ Split into chunks of x's length. | [[['a', 'b', 'c', '1', '2', 'D', '5']], [['a', 'b', 'c', '1', '2', 'D', '5'], ['a', 'b', 'c', '1', '2', 'D', '5'], ['a', 'b', 'c', '1']], [['a', 'b', 'C', '3', '4', 'd', '3'], ['a', 'b', 'C', '3', '4', 'd', '3'], ['a', 'b', 'C', '3']]]
          ŒsÐe Swap the case of even-indexed chunks (1-indexed). | [[['a', 'b', 'c', '1', '2', 'D', '5']], [['a', 'b', 'c', '1', '2', 'D', '5'], ['A', 'B', 'C', '1', '2', 'd', '5'], ['a', 'b', 'c', '1']], [['a', 'b', 'C', '3', '4', 'd', '3'], ['A', 'B', 'c', '3', '4', 'D', '3'], ['a', 'b', 'C', '3']]]
          F Flatten. | [['a', 'b', 'c', '1', '2', 'D', '5'], ['a', 'b', 'c', '1', '2', 'D', '5', 'A', 'B', 'C', '1', '2', 'd', '5', 'a', 'b', 'c', '1'], ['a', 'b', 'C', '3', '4', 'd', '3', 'A', 'B', 'c', '3', '4', 'D', '3', 'a', 'b', 'C', '3']]
          n Vectorized inequality with y. | [[[0, 0, 1, 1, 1, 1, 1]], [[1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]], [[1, 1, 0, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1]]]
          § After ending the map, sum each bool (0 or 1) array. | [[5], [17], [14]]
          Ṁ Maximum. | 17





          share|improve this answer






















          • I'm totally not familiar with Jelly, but I think you can omit LÞ and still get the same maximum at the end.
            – Chas Brown
            Aug 13 at 21:41











          • @ChasBrown Ugh, no, I should need that. Otherwise, instead of padding the shortest to the length of the longest one, ṁ/ would instead trim the longest to the length of the shortest one in some cases, which is not what we want.... I guess the test cases are too well chosen (and this is a rather unfortunate coincidence)...
            – Mr. Xcoder
            Aug 13 at 21:48











          • @ChasBrown As an example, try ["AacaAc", "Aab"].
            – Mr. Xcoder
            Aug 13 at 21:57










          • Ah yes, I see... I need to learn me some more Jelly... :)
            – Chas Brown
            Aug 13 at 22:02














          up vote
          7
          down vote














          Jelly, 20 bytes



          Not really happy with it. Should be golfable, even to ~15 bytes perhaps.



          LÞŒcµṁ/sḢL$ŒsÐeFn)§Ṁ


          Try it online!



          or Check out a test suite!



          Explanation




          LÞŒcµṁ/sḢL$ŒsÐeFn)§Ṁ Full program or monadic link. N = input. | Example: ["abc12D5", "abC34d3", "ABC14dabc23DAbC89d"]
          LÞ Sort N by length. | [['a', 'b', 'c', '1', '2', 'D', '5'], ['a', 'b', 'C', '3', '4', 'd', '3'], ['A', 'B', 'C', '1', '4', 'd', 'a', 'b', 'c', '2', '3', 'D', 'A', 'b', 'C', '8', '9', 'd']] (in Jelly, strings are list of characters)
          Ã…Â’c Unordered pairs: [x, y] for all distinct x, y in N. | [[['a', 'b', 'c', '1', '2', 'D', '5'], ['a', 'b', 'C', '3', '4', 'd', '3']], [['a', 'b', 'c', '1', '2', 'D', '5'], ['A', 'B', 'C', '1', '4', 'd', 'a', 'b', 'c', '2', '3', 'D', 'A', 'b', 'C', '8', '9', 'd']], [['a', 'b', 'C', '3', '4', 'd', '3'], ['A', 'B', 'C', '1', '4', 'd', 'a', 'b', 'c', '2', '3', 'D', 'A', 'b', 'C', '8', '9', 'd']]]
          Here, by distinct, I mean at different positions. |
          µ ) Map with a monadic link. |
          ṁ/ Mold x like y. That is, cycle x until it reaches length y. | [['a', 'b', 'c', '1', '2', 'D', '5'], ['a', 'b', 'c', '1', '2', 'D', '5', 'a', 'b', 'c', '1', '2', 'D', '5', 'a', 'b', 'c', '1'], ['a', 'b', 'C', '3', '4', 'd', '3', 'a', 'b', 'C', '3', '4', 'd', '3', 'a', 'b', 'C', '3']]
          sḢL$ Split into chunks of x's length. | [[['a', 'b', 'c', '1', '2', 'D', '5']], [['a', 'b', 'c', '1', '2', 'D', '5'], ['a', 'b', 'c', '1', '2', 'D', '5'], ['a', 'b', 'c', '1']], [['a', 'b', 'C', '3', '4', 'd', '3'], ['a', 'b', 'C', '3', '4', 'd', '3'], ['a', 'b', 'C', '3']]]
          ŒsÐe Swap the case of even-indexed chunks (1-indexed). | [[['a', 'b', 'c', '1', '2', 'D', '5']], [['a', 'b', 'c', '1', '2', 'D', '5'], ['A', 'B', 'C', '1', '2', 'd', '5'], ['a', 'b', 'c', '1']], [['a', 'b', 'C', '3', '4', 'd', '3'], ['A', 'B', 'c', '3', '4', 'D', '3'], ['a', 'b', 'C', '3']]]
          F Flatten. | [['a', 'b', 'c', '1', '2', 'D', '5'], ['a', 'b', 'c', '1', '2', 'D', '5', 'A', 'B', 'C', '1', '2', 'd', '5', 'a', 'b', 'c', '1'], ['a', 'b', 'C', '3', '4', 'd', '3', 'A', 'B', 'c', '3', '4', 'D', '3', 'a', 'b', 'C', '3']]
          n Vectorized inequality with y. | [[[0, 0, 1, 1, 1, 1, 1]], [[1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]], [[1, 1, 0, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1]]]
          § After ending the map, sum each bool (0 or 1) array. | [[5], [17], [14]]
          Ṁ Maximum. | 17





          share|improve this answer






















          • I'm totally not familiar with Jelly, but I think you can omit LÞ and still get the same maximum at the end.
            – Chas Brown
            Aug 13 at 21:41











          • @ChasBrown Ugh, no, I should need that. Otherwise, instead of padding the shortest to the length of the longest one, ṁ/ would instead trim the longest to the length of the shortest one in some cases, which is not what we want.... I guess the test cases are too well chosen (and this is a rather unfortunate coincidence)...
            – Mr. Xcoder
            Aug 13 at 21:48











          • @ChasBrown As an example, try ["AacaAc", "Aab"].
            – Mr. Xcoder
            Aug 13 at 21:57










          • Ah yes, I see... I need to learn me some more Jelly... :)
            – Chas Brown
            Aug 13 at 22:02












          up vote
          7
          down vote










          up vote
          7
          down vote










          Jelly, 20 bytes



          Not really happy with it. Should be golfable, even to ~15 bytes perhaps.



          LÞŒcµṁ/sḢL$ŒsÐeFn)§Ṁ


          Try it online!



          or Check out a test suite!



          Explanation




          LÞŒcµṁ/sḢL$ŒsÐeFn)§Ṁ Full program or monadic link. N = input. | Example: ["abc12D5", "abC34d3", "ABC14dabc23DAbC89d"]
          LÞ Sort N by length. | [['a', 'b', 'c', '1', '2', 'D', '5'], ['a', 'b', 'C', '3', '4', 'd', '3'], ['A', 'B', 'C', '1', '4', 'd', 'a', 'b', 'c', '2', '3', 'D', 'A', 'b', 'C', '8', '9', 'd']] (in Jelly, strings are list of characters)
          Ã…Â’c Unordered pairs: [x, y] for all distinct x, y in N. | [[['a', 'b', 'c', '1', '2', 'D', '5'], ['a', 'b', 'C', '3', '4', 'd', '3']], [['a', 'b', 'c', '1', '2', 'D', '5'], ['A', 'B', 'C', '1', '4', 'd', 'a', 'b', 'c', '2', '3', 'D', 'A', 'b', 'C', '8', '9', 'd']], [['a', 'b', 'C', '3', '4', 'd', '3'], ['A', 'B', 'C', '1', '4', 'd', 'a', 'b', 'c', '2', '3', 'D', 'A', 'b', 'C', '8', '9', 'd']]]
          Here, by distinct, I mean at different positions. |
          µ ) Map with a monadic link. |
          ṁ/ Mold x like y. That is, cycle x until it reaches length y. | [['a', 'b', 'c', '1', '2', 'D', '5'], ['a', 'b', 'c', '1', '2', 'D', '5', 'a', 'b', 'c', '1', '2', 'D', '5', 'a', 'b', 'c', '1'], ['a', 'b', 'C', '3', '4', 'd', '3', 'a', 'b', 'C', '3', '4', 'd', '3', 'a', 'b', 'C', '3']]
          sḢL$ Split into chunks of x's length. | [[['a', 'b', 'c', '1', '2', 'D', '5']], [['a', 'b', 'c', '1', '2', 'D', '5'], ['a', 'b', 'c', '1', '2', 'D', '5'], ['a', 'b', 'c', '1']], [['a', 'b', 'C', '3', '4', 'd', '3'], ['a', 'b', 'C', '3', '4', 'd', '3'], ['a', 'b', 'C', '3']]]
          ŒsÐe Swap the case of even-indexed chunks (1-indexed). | [[['a', 'b', 'c', '1', '2', 'D', '5']], [['a', 'b', 'c', '1', '2', 'D', '5'], ['A', 'B', 'C', '1', '2', 'd', '5'], ['a', 'b', 'c', '1']], [['a', 'b', 'C', '3', '4', 'd', '3'], ['A', 'B', 'c', '3', '4', 'D', '3'], ['a', 'b', 'C', '3']]]
          F Flatten. | [['a', 'b', 'c', '1', '2', 'D', '5'], ['a', 'b', 'c', '1', '2', 'D', '5', 'A', 'B', 'C', '1', '2', 'd', '5', 'a', 'b', 'c', '1'], ['a', 'b', 'C', '3', '4', 'd', '3', 'A', 'B', 'c', '3', '4', 'D', '3', 'a', 'b', 'C', '3']]
          n Vectorized inequality with y. | [[[0, 0, 1, 1, 1, 1, 1]], [[1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]], [[1, 1, 0, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1]]]
          § After ending the map, sum each bool (0 or 1) array. | [[5], [17], [14]]
          Ṁ Maximum. | 17





          share|improve this answer















          Jelly, 20 bytes



          Not really happy with it. Should be golfable, even to ~15 bytes perhaps.



          LÞŒcµṁ/sḢL$ŒsÐeFn)§Ṁ


          Try it online!



          or Check out a test suite!



          Explanation




          LÞŒcµṁ/sḢL$ŒsÐeFn)§Ṁ Full program or monadic link. N = input. | Example: ["abc12D5", "abC34d3", "ABC14dabc23DAbC89d"]
          LÞ Sort N by length. | [['a', 'b', 'c', '1', '2', 'D', '5'], ['a', 'b', 'C', '3', '4', 'd', '3'], ['A', 'B', 'C', '1', '4', 'd', 'a', 'b', 'c', '2', '3', 'D', 'A', 'b', 'C', '8', '9', 'd']] (in Jelly, strings are list of characters)
          Ã…Â’c Unordered pairs: [x, y] for all distinct x, y in N. | [[['a', 'b', 'c', '1', '2', 'D', '5'], ['a', 'b', 'C', '3', '4', 'd', '3']], [['a', 'b', 'c', '1', '2', 'D', '5'], ['A', 'B', 'C', '1', '4', 'd', 'a', 'b', 'c', '2', '3', 'D', 'A', 'b', 'C', '8', '9', 'd']], [['a', 'b', 'C', '3', '4', 'd', '3'], ['A', 'B', 'C', '1', '4', 'd', 'a', 'b', 'c', '2', '3', 'D', 'A', 'b', 'C', '8', '9', 'd']]]
          Here, by distinct, I mean at different positions. |
          µ ) Map with a monadic link. |
          ṁ/ Mold x like y. That is, cycle x until it reaches length y. | [['a', 'b', 'c', '1', '2', 'D', '5'], ['a', 'b', 'c', '1', '2', 'D', '5', 'a', 'b', 'c', '1', '2', 'D', '5', 'a', 'b', 'c', '1'], ['a', 'b', 'C', '3', '4', 'd', '3', 'a', 'b', 'C', '3', '4', 'd', '3', 'a', 'b', 'C', '3']]
          sḢL$ Split into chunks of x's length. | [[['a', 'b', 'c', '1', '2', 'D', '5']], [['a', 'b', 'c', '1', '2', 'D', '5'], ['a', 'b', 'c', '1', '2', 'D', '5'], ['a', 'b', 'c', '1']], [['a', 'b', 'C', '3', '4', 'd', '3'], ['a', 'b', 'C', '3', '4', 'd', '3'], ['a', 'b', 'C', '3']]]
          ŒsÐe Swap the case of even-indexed chunks (1-indexed). | [[['a', 'b', 'c', '1', '2', 'D', '5']], [['a', 'b', 'c', '1', '2', 'D', '5'], ['A', 'B', 'C', '1', '2', 'd', '5'], ['a', 'b', 'c', '1']], [['a', 'b', 'C', '3', '4', 'd', '3'], ['A', 'B', 'c', '3', '4', 'D', '3'], ['a', 'b', 'C', '3']]]
          F Flatten. | [['a', 'b', 'c', '1', '2', 'D', '5'], ['a', 'b', 'c', '1', '2', 'D', '5', 'A', 'B', 'C', '1', '2', 'd', '5', 'a', 'b', 'c', '1'], ['a', 'b', 'C', '3', '4', 'd', '3', 'A', 'B', 'c', '3', '4', 'D', '3', 'a', 'b', 'C', '3']]
          n Vectorized inequality with y. | [[[0, 0, 1, 1, 1, 1, 1]], [[1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]], [[1, 1, 0, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1]]]
          § After ending the map, sum each bool (0 or 1) array. | [[5], [17], [14]]
          Ṁ Maximum. | 17






          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited Aug 13 at 20:32

























          answered Aug 13 at 20:03









          Mr. Xcoder

          30.2k757193




          30.2k757193











          • I'm totally not familiar with Jelly, but I think you can omit LÞ and still get the same maximum at the end.
            – Chas Brown
            Aug 13 at 21:41











          • @ChasBrown Ugh, no, I should need that. Otherwise, instead of padding the shortest to the length of the longest one, ṁ/ would instead trim the longest to the length of the shortest one in some cases, which is not what we want.... I guess the test cases are too well chosen (and this is a rather unfortunate coincidence)...
            – Mr. Xcoder
            Aug 13 at 21:48











          • @ChasBrown As an example, try ["AacaAc", "Aab"].
            – Mr. Xcoder
            Aug 13 at 21:57










          • Ah yes, I see... I need to learn me some more Jelly... :)
            – Chas Brown
            Aug 13 at 22:02
















          • I'm totally not familiar with Jelly, but I think you can omit LÞ and still get the same maximum at the end.
            – Chas Brown
            Aug 13 at 21:41











          • @ChasBrown Ugh, no, I should need that. Otherwise, instead of padding the shortest to the length of the longest one, ṁ/ would instead trim the longest to the length of the shortest one in some cases, which is not what we want.... I guess the test cases are too well chosen (and this is a rather unfortunate coincidence)...
            – Mr. Xcoder
            Aug 13 at 21:48











          • @ChasBrown As an example, try ["AacaAc", "Aab"].
            – Mr. Xcoder
            Aug 13 at 21:57










          • Ah yes, I see... I need to learn me some more Jelly... :)
            – Chas Brown
            Aug 13 at 22:02















          I'm totally not familiar with Jelly, but I think you can omit LÞ and still get the same maximum at the end.
          – Chas Brown
          Aug 13 at 21:41





          I'm totally not familiar with Jelly, but I think you can omit LÞ and still get the same maximum at the end.
          – Chas Brown
          Aug 13 at 21:41













          @ChasBrown Ugh, no, I should need that. Otherwise, instead of padding the shortest to the length of the longest one, ṁ/ would instead trim the longest to the length of the shortest one in some cases, which is not what we want.... I guess the test cases are too well chosen (and this is a rather unfortunate coincidence)...
          – Mr. Xcoder
          Aug 13 at 21:48





          @ChasBrown Ugh, no, I should need that. Otherwise, instead of padding the shortest to the length of the longest one, ṁ/ would instead trim the longest to the length of the shortest one in some cases, which is not what we want.... I guess the test cases are too well chosen (and this is a rather unfortunate coincidence)...
          – Mr. Xcoder
          Aug 13 at 21:48













          @ChasBrown As an example, try ["AacaAc", "Aab"].
          – Mr. Xcoder
          Aug 13 at 21:57




          @ChasBrown As an example, try ["AacaAc", "Aab"].
          – Mr. Xcoder
          Aug 13 at 21:57












          Ah yes, I see... I need to learn me some more Jelly... :)
          – Chas Brown
          Aug 13 at 22:02




          Ah yes, I see... I need to learn me some more Jelly... :)
          – Chas Brown
          Aug 13 at 22:02










          up vote
          5
          down vote














          Python 2, 86 bytes





          lambda a:max(sum(x!=y for x,y in zip((s+s.swapcase())*len(t),t))for s in a for t in a)


          Try it online!



          Given two strings, s,t, zip((s+s.swapcase())*len(t),t)) will be a list of tuples of length len(t) since zip truncates to the shortest iterable. If len(s)<len(t), then this "pads out" s with the desired case swapping and we calculate the sum of differing characters.



          If len(t)<=len(s), then the resulting sum will be less than or equal to the sum if we were evaluating t,s; so it has no effect on the resulting max in that case.






          share|improve this answer






















          • You can use y!= instead of !=y to save 1 byte
            – Mr. Xcoder
            Aug 13 at 20:43










          • @Mr. Xcoder: Thx, but I ended up drastically reworking my solution...
            – Chas Brown
            Aug 13 at 21:10














          up vote
          5
          down vote














          Python 2, 86 bytes





          lambda a:max(sum(x!=y for x,y in zip((s+s.swapcase())*len(t),t))for s in a for t in a)


          Try it online!



          Given two strings, s,t, zip((s+s.swapcase())*len(t),t)) will be a list of tuples of length len(t) since zip truncates to the shortest iterable. If len(s)<len(t), then this "pads out" s with the desired case swapping and we calculate the sum of differing characters.



          If len(t)<=len(s), then the resulting sum will be less than or equal to the sum if we were evaluating t,s; so it has no effect on the resulting max in that case.






          share|improve this answer






















          • You can use y!= instead of !=y to save 1 byte
            – Mr. Xcoder
            Aug 13 at 20:43










          • @Mr. Xcoder: Thx, but I ended up drastically reworking my solution...
            – Chas Brown
            Aug 13 at 21:10












          up vote
          5
          down vote










          up vote
          5
          down vote










          Python 2, 86 bytes





          lambda a:max(sum(x!=y for x,y in zip((s+s.swapcase())*len(t),t))for s in a for t in a)


          Try it online!



          Given two strings, s,t, zip((s+s.swapcase())*len(t),t)) will be a list of tuples of length len(t) since zip truncates to the shortest iterable. If len(s)<len(t), then this "pads out" s with the desired case swapping and we calculate the sum of differing characters.



          If len(t)<=len(s), then the resulting sum will be less than or equal to the sum if we were evaluating t,s; so it has no effect on the resulting max in that case.






          share|improve this answer















          Python 2, 86 bytes





          lambda a:max(sum(x!=y for x,y in zip((s+s.swapcase())*len(t),t))for s in a for t in a)


          Try it online!



          Given two strings, s,t, zip((s+s.swapcase())*len(t),t)) will be a list of tuples of length len(t) since zip truncates to the shortest iterable. If len(s)<len(t), then this "pads out" s with the desired case swapping and we calculate the sum of differing characters.



          If len(t)<=len(s), then the resulting sum will be less than or equal to the sum if we were evaluating t,s; so it has no effect on the resulting max in that case.







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited Aug 13 at 21:07

























          answered Aug 13 at 20:38









          Chas Brown

          4,1361319




          4,1361319











          • You can use y!= instead of !=y to save 1 byte
            – Mr. Xcoder
            Aug 13 at 20:43










          • @Mr. Xcoder: Thx, but I ended up drastically reworking my solution...
            – Chas Brown
            Aug 13 at 21:10
















          • You can use y!= instead of !=y to save 1 byte
            – Mr. Xcoder
            Aug 13 at 20:43










          • @Mr. Xcoder: Thx, but I ended up drastically reworking my solution...
            – Chas Brown
            Aug 13 at 21:10















          You can use y!= instead of !=y to save 1 byte
          – Mr. Xcoder
          Aug 13 at 20:43




          You can use y!= instead of !=y to save 1 byte
          – Mr. Xcoder
          Aug 13 at 20:43












          @Mr. Xcoder: Thx, but I ended up drastically reworking my solution...
          – Chas Brown
          Aug 13 at 21:10




          @Mr. Xcoder: Thx, but I ended up drastically reworking my solution...
          – Chas Brown
          Aug 13 at 21:10










          up vote
          3
          down vote














          JavaScript (Node.js), 111 bytes





          a=>a.map(m=S=>a.map(s=>B(S).map((c,k)=>m=(c^(c=B(s)[k%(l=s.length)])^(k/l&c>9)<<5&&++x)<m?m:x,x=0)),B=Buffer)|m


          Try it online!






          share|improve this answer
























            up vote
            3
            down vote














            JavaScript (Node.js), 111 bytes





            a=>a.map(m=S=>a.map(s=>B(S).map((c,k)=>m=(c^(c=B(s)[k%(l=s.length)])^(k/l&c>9)<<5&&++x)<m?m:x,x=0)),B=Buffer)|m


            Try it online!






            share|improve this answer






















              up vote
              3
              down vote










              up vote
              3
              down vote










              JavaScript (Node.js), 111 bytes





              a=>a.map(m=S=>a.map(s=>B(S).map((c,k)=>m=(c^(c=B(s)[k%(l=s.length)])^(k/l&c>9)<<5&&++x)<m?m:x,x=0)),B=Buffer)|m


              Try it online!






              share|improve this answer













              JavaScript (Node.js), 111 bytes





              a=>a.map(m=S=>a.map(s=>B(S).map((c,k)=>m=(c^(c=B(s)[k%(l=s.length)])^(k/l&c>9)<<5&&++x)<m?m:x,x=0)),B=Buffer)|m


              Try it online!







              share|improve this answer












              share|improve this answer



              share|improve this answer










              answered Aug 13 at 23:24









              Arnauld

              63.3k580267




              63.3k580267




















                  up vote
                  3
                  down vote














                  Jelly, 19 bytes



                  WṁŒsÐeF=ċ0
                  LÞŒcç/€Ṁ


                  Try it online!



                  LÞŒcç/€Ṁ
                  LÞ Sort by length
                  Ã…Â’c unordered pairs
                  € to each of the pairs
                  ç/ find the hamming distance with molding and swapping case (helper link)
                  Ṁ maximum

                  WṁŒsÐeF=ċ0
                  W wrap the shorter string
                  ṁ repeat this string once for each character in the second string
                  Ðe for every other repeated string
                  Ã…Â’s swap case
                  F flatten
                  = characterwise equality check between the two strings. If the first
                  string is longer, the leftover characters are appended to the result.
                  e.g. 'abABab' and 'xbA' give [0,1,1,'B','a','b']
                  ċ0 count the number of 0s, giving the Hamming distance.





                  share|improve this answer


























                    up vote
                    3
                    down vote














                    Jelly, 19 bytes



                    WṁŒsÐeF=ċ0
                    LÞŒcç/€Ṁ


                    Try it online!



                    LÞŒcç/€Ṁ
                    LÞ Sort by length
                    Ã…Â’c unordered pairs
                    € to each of the pairs
                    ç/ find the hamming distance with molding and swapping case (helper link)
                    Ṁ maximum

                    WṁŒsÐeF=ċ0
                    W wrap the shorter string
                    ṁ repeat this string once for each character in the second string
                    Ðe for every other repeated string
                    Ã…Â’s swap case
                    F flatten
                    = characterwise equality check between the two strings. If the first
                    string is longer, the leftover characters are appended to the result.
                    e.g. 'abABab' and 'xbA' give [0,1,1,'B','a','b']
                    ċ0 count the number of 0s, giving the Hamming distance.





                    share|improve this answer
























                      up vote
                      3
                      down vote










                      up vote
                      3
                      down vote










                      Jelly, 19 bytes



                      WṁŒsÐeF=ċ0
                      LÞŒcç/€Ṁ


                      Try it online!



                      LÞŒcç/€Ṁ
                      LÞ Sort by length
                      Ã…Â’c unordered pairs
                      € to each of the pairs
                      ç/ find the hamming distance with molding and swapping case (helper link)
                      Ṁ maximum

                      WṁŒsÐeF=ċ0
                      W wrap the shorter string
                      ṁ repeat this string once for each character in the second string
                      Ðe for every other repeated string
                      Ã…Â’s swap case
                      F flatten
                      = characterwise equality check between the two strings. If the first
                      string is longer, the leftover characters are appended to the result.
                      e.g. 'abABab' and 'xbA' give [0,1,1,'B','a','b']
                      ċ0 count the number of 0s, giving the Hamming distance.





                      share|improve this answer















                      Jelly, 19 bytes



                      WṁŒsÐeF=ċ0
                      LÞŒcç/€Ṁ


                      Try it online!



                      LÞŒcç/€Ṁ
                      LÞ Sort by length
                      Ã…Â’c unordered pairs
                      € to each of the pairs
                      ç/ find the hamming distance with molding and swapping case (helper link)
                      Ṁ maximum

                      WṁŒsÐeF=ċ0
                      W wrap the shorter string
                      ṁ repeat this string once for each character in the second string
                      Ðe for every other repeated string
                      Ã…Â’s swap case
                      F flatten
                      = characterwise equality check between the two strings. If the first
                      string is longer, the leftover characters are appended to the result.
                      e.g. 'abABab' and 'xbA' give [0,1,1,'B','a','b']
                      ċ0 count the number of 0s, giving the Hamming distance.






                      share|improve this answer














                      share|improve this answer



                      share|improve this answer








                      edited Aug 14 at 17:48

























                      answered Aug 14 at 17:41









                      dylnan

                      3,4982526




                      3,4982526




















                          up vote
                          2
                          down vote














                          Ruby, 89 82 bytes



                          Creates the cross-product of the input list against itself before calculating the Hamming distance of each pair, using a duplication method similar to Chas Brown's answer. Ruby can't zip strings together or add booleans without additional overhead, though, so it becomes necessary to iterate through the pair of strings manually instead.



                          -7 bytes from GB.





                          ->aa.product(a).maps,t.max


                          Try it online!






                          share|improve this answer


















                          • 1




                            82 bytes
                            – G B
                            Aug 14 at 11:27














                          up vote
                          2
                          down vote














                          Ruby, 89 82 bytes



                          Creates the cross-product of the input list against itself before calculating the Hamming distance of each pair, using a duplication method similar to Chas Brown's answer. Ruby can't zip strings together or add booleans without additional overhead, though, so it becomes necessary to iterate through the pair of strings manually instead.



                          -7 bytes from GB.





                          ->aa.product(a).maps,t.max


                          Try it online!






                          share|improve this answer


















                          • 1




                            82 bytes
                            – G B
                            Aug 14 at 11:27












                          up vote
                          2
                          down vote










                          up vote
                          2
                          down vote










                          Ruby, 89 82 bytes



                          Creates the cross-product of the input list against itself before calculating the Hamming distance of each pair, using a duplication method similar to Chas Brown's answer. Ruby can't zip strings together or add booleans without additional overhead, though, so it becomes necessary to iterate through the pair of strings manually instead.



                          -7 bytes from GB.





                          ->aa.product(a).maps,t.max


                          Try it online!






                          share|improve this answer















                          Ruby, 89 82 bytes



                          Creates the cross-product of the input list against itself before calculating the Hamming distance of each pair, using a duplication method similar to Chas Brown's answer. Ruby can't zip strings together or add booleans without additional overhead, though, so it becomes necessary to iterate through the pair of strings manually instead.



                          -7 bytes from GB.





                          ->aa.product(a).maps,t.max


                          Try it online!







                          share|improve this answer














                          share|improve this answer



                          share|improve this answer








                          edited Aug 14 at 22:59

























                          answered Aug 14 at 3:31









                          Value Ink

                          7,827630




                          7,827630







                          • 1




                            82 bytes
                            – G B
                            Aug 14 at 11:27












                          • 1




                            82 bytes
                            – G B
                            Aug 14 at 11:27







                          1




                          1




                          82 bytes
                          – G B
                          Aug 14 at 11:27




                          82 bytes
                          – G B
                          Aug 14 at 11:27










                          up vote
                          2
                          down vote














                          Java 10, 748 740 667 666 616 bytes



                          This has to be the most dense and unreadable, yet the longest golf I ever came up with.



                          Call method h(String) with an explicit array (no var args): eg,



                          h(new String "a", "b", "c");


                          returns 1.



                          char e(boolean w,char c)96<c&c<123)?c^32:c);String p(String s,int l)var p="";int m=s.length(),n=l/m,r=l%m,i=0,j=0;var w=1<0;for(;i<n;++i,w=!w)for(char c:s.toCharArray())p+=e(w,c);for(;j<r;)p+=e(w,s.charAt(j++));return p;int d(String s,String t)int l=s.length(),n=0,i=0;for(;i<l;)if(s.charAt(i)!=t.charAt(i++))++n;return n;int h(String s,String t)int l=s.length(),m=t.length();return l>m?d(s,p(t,l)):l<m?d(p(s,m),t):d(s,t);int h(Strings)int l=s.length,i=0,j;intn=new int[l*l];for(;i<l;++i)for(j=i;++j<l;)n[i*l+j]=h(s[i],s[j]);return java.util.Arrays.stream(n).max().getAsInt();


                          You can Try It Online!



                          Ungolfed and commented:



                          // Encode the character (swap case)
                          char e(boolean w, char c) 96 < c & c < 123) ? c ^ 32 : c);


                          // Pad the string to desired length
                          String p(String s, int l)
                          var p = "";
                          int m = s.length(), n = l / m, r = l % m, i = 0, j = 0;
                          var w = 1 < 0;
                          for (; i < n; ++i, w = !w)
                          for (char c : s.toCharArray())
                          p += e(w, c);
                          for (; j < r;)
                          p += e(w, s.charAt(j++));
                          return p;


                          // Calculate the actual hamming distance between two same-length strings
                          int d(String s, String t)
                          int l = s.length(), n = 0, i = 0;
                          for (; i < l;)
                          if (s.charAt(i) != t.charAt(i++))
                          ++n;
                          return n;

                          // Pad the strings as needed and return their hamming distance
                          int h(String s, String t)
                          int l = s.length(), m = t.length();
                          return l > m ? d(s, p(t, l)) : l < m ? d(p(s, m), t) : d(s, t);


                          // Dispatch the strings and gather their hamming distances, return the max
                          int h(String s)
                          int l = s.length, i = 0, j;
                          int n = new int[l * l];
                          for (; i < l; ++i)
                          for (j = i; ++j < l;)
                          n[i * l + j] = h(s[i], s[j]);
                          return java.util.Arrays.stream(n).max().getAsInt();



                          I know a better solution can be achieved, especially for the string pairing part.



                          EDIT: shave off 8 bytes by changing the size of the int array in hammingDistance() to the square of the numbe of strings given. It also fixes an ArrayIndexOutOfBounds thrown in one of the test cases.



                          EDIT 2: Saved 33 bytes thanks to Kevin Cruijssen's comments: class declaration removed, names shortened to 1 char, operators changed, etc.



                          EDIT 3: Save 1 byte and reach Satan-approved score by changing method with var-arg to array.



                          EDIT 4: Save another 50 bytes thanks to Kevin Cruijssen, again: update Java version from 8 to 10 to use var keyword, removed StringBuilder instance, etc.






                          share|improve this answer


















                          • 1




                            I don't have a lot of time, but some basic things to golf: Drop the class, only the methods is enough. Change all method and variable names to single bytes.. So instead of hammingDistance use d or some other unused variable. Most of your && can be & and || can be |. c^' ' can be c^32. boolean w = false; can be boolean w=0>1;. i=0 in the loop initialization can be removed and change the ,i,j to ,i=0,j. ++j can be removed and ++ can be added to the .charAt(j++). .toString() can be +"". for(j=i+1;j<l;++j) can be for(j=0;++j<l;). Etc. etc.
                            – Kevin Cruijssen
                            Aug 14 at 13:38






                          • 1




                            Tips for golfing in Java and Tips for golfing in <all languages> might be interesting to read through as well. :)
                            – Kevin Cruijssen
                            Aug 14 at 13:41










                          • Thanks! That's some nice byte-lifting. Thanks for the links too, I'm taking a look at it and will edit asap!
                            – joH1
                            Aug 14 at 13:47






                          • 1




                            Upvoted for the Satan-approved score. xD Some more small things: StringBuilder can be StringBuffer (if you switch to Java 10 it could be var b=new StringBuffer(l);. The boolean and char can then also be var. If you don't have Java 10 locally, it is available on TIO). In addition, for(;i<n;++i)for(char c:s.toCharArray())b.append(e(w,c));w=!w; can be for(;i++<n;w=!w)for(char c:s.toCharArray())b.append(e(w,c));. And I'm pretty sure you can remove the StringBuffer completely and just use String and += instead of append.
                            – Kevin Cruijssen
                            Aug 14 at 15:13











                          • Man, some many months of clean code and good coding practices made me forget how to even golf! I'll update my answer and include TIO.
                            – joH1
                            Aug 16 at 7:15














                          up vote
                          2
                          down vote














                          Java 10, 748 740 667 666 616 bytes



                          This has to be the most dense and unreadable, yet the longest golf I ever came up with.



                          Call method h(String) with an explicit array (no var args): eg,



                          h(new String "a", "b", "c");


                          returns 1.



                          char e(boolean w,char c)96<c&c<123)?c^32:c);String p(String s,int l)var p="";int m=s.length(),n=l/m,r=l%m,i=0,j=0;var w=1<0;for(;i<n;++i,w=!w)for(char c:s.toCharArray())p+=e(w,c);for(;j<r;)p+=e(w,s.charAt(j++));return p;int d(String s,String t)int l=s.length(),n=0,i=0;for(;i<l;)if(s.charAt(i)!=t.charAt(i++))++n;return n;int h(String s,String t)int l=s.length(),m=t.length();return l>m?d(s,p(t,l)):l<m?d(p(s,m),t):d(s,t);int h(Strings)int l=s.length,i=0,j;intn=new int[l*l];for(;i<l;++i)for(j=i;++j<l;)n[i*l+j]=h(s[i],s[j]);return java.util.Arrays.stream(n).max().getAsInt();


                          You can Try It Online!



                          Ungolfed and commented:



                          // Encode the character (swap case)
                          char e(boolean w, char c) 96 < c & c < 123) ? c ^ 32 : c);


                          // Pad the string to desired length
                          String p(String s, int l)
                          var p = "";
                          int m = s.length(), n = l / m, r = l % m, i = 0, j = 0;
                          var w = 1 < 0;
                          for (; i < n; ++i, w = !w)
                          for (char c : s.toCharArray())
                          p += e(w, c);
                          for (; j < r;)
                          p += e(w, s.charAt(j++));
                          return p;


                          // Calculate the actual hamming distance between two same-length strings
                          int d(String s, String t)
                          int l = s.length(), n = 0, i = 0;
                          for (; i < l;)
                          if (s.charAt(i) != t.charAt(i++))
                          ++n;
                          return n;

                          // Pad the strings as needed and return their hamming distance
                          int h(String s, String t)
                          int l = s.length(), m = t.length();
                          return l > m ? d(s, p(t, l)) : l < m ? d(p(s, m), t) : d(s, t);


                          // Dispatch the strings and gather their hamming distances, return the max
                          int h(String s)
                          int l = s.length, i = 0, j;
                          int n = new int[l * l];
                          for (; i < l; ++i)
                          for (j = i; ++j < l;)
                          n[i * l + j] = h(s[i], s[j]);
                          return java.util.Arrays.stream(n).max().getAsInt();



                          I know a better solution can be achieved, especially for the string pairing part.



                          EDIT: shave off 8 bytes by changing the size of the int array in hammingDistance() to the square of the numbe of strings given. It also fixes an ArrayIndexOutOfBounds thrown in one of the test cases.



                          EDIT 2: Saved 33 bytes thanks to Kevin Cruijssen's comments: class declaration removed, names shortened to 1 char, operators changed, etc.



                          EDIT 3: Save 1 byte and reach Satan-approved score by changing method with var-arg to array.



                          EDIT 4: Save another 50 bytes thanks to Kevin Cruijssen, again: update Java version from 8 to 10 to use var keyword, removed StringBuilder instance, etc.






                          share|improve this answer


















                          • 1




                            I don't have a lot of time, but some basic things to golf: Drop the class, only the methods is enough. Change all method and variable names to single bytes.. So instead of hammingDistance use d or some other unused variable. Most of your && can be & and || can be |. c^' ' can be c^32. boolean w = false; can be boolean w=0>1;. i=0 in the loop initialization can be removed and change the ,i,j to ,i=0,j. ++j can be removed and ++ can be added to the .charAt(j++). .toString() can be +"". for(j=i+1;j<l;++j) can be for(j=0;++j<l;). Etc. etc.
                            – Kevin Cruijssen
                            Aug 14 at 13:38






                          • 1




                            Tips for golfing in Java and Tips for golfing in <all languages> might be interesting to read through as well. :)
                            – Kevin Cruijssen
                            Aug 14 at 13:41










                          • Thanks! That's some nice byte-lifting. Thanks for the links too, I'm taking a look at it and will edit asap!
                            – joH1
                            Aug 14 at 13:47






                          • 1




                            Upvoted for the Satan-approved score. xD Some more small things: StringBuilder can be StringBuffer (if you switch to Java 10 it could be var b=new StringBuffer(l);. The boolean and char can then also be var. If you don't have Java 10 locally, it is available on TIO). In addition, for(;i<n;++i)for(char c:s.toCharArray())b.append(e(w,c));w=!w; can be for(;i++<n;w=!w)for(char c:s.toCharArray())b.append(e(w,c));. And I'm pretty sure you can remove the StringBuffer completely and just use String and += instead of append.
                            – Kevin Cruijssen
                            Aug 14 at 15:13











                          • Man, some many months of clean code and good coding practices made me forget how to even golf! I'll update my answer and include TIO.
                            – joH1
                            Aug 16 at 7:15












                          up vote
                          2
                          down vote










                          up vote
                          2
                          down vote










                          Java 10, 748 740 667 666 616 bytes



                          This has to be the most dense and unreadable, yet the longest golf I ever came up with.



                          Call method h(String) with an explicit array (no var args): eg,



                          h(new String "a", "b", "c");


                          returns 1.



                          char e(boolean w,char c)96<c&c<123)?c^32:c);String p(String s,int l)var p="";int m=s.length(),n=l/m,r=l%m,i=0,j=0;var w=1<0;for(;i<n;++i,w=!w)for(char c:s.toCharArray())p+=e(w,c);for(;j<r;)p+=e(w,s.charAt(j++));return p;int d(String s,String t)int l=s.length(),n=0,i=0;for(;i<l;)if(s.charAt(i)!=t.charAt(i++))++n;return n;int h(String s,String t)int l=s.length(),m=t.length();return l>m?d(s,p(t,l)):l<m?d(p(s,m),t):d(s,t);int h(Strings)int l=s.length,i=0,j;intn=new int[l*l];for(;i<l;++i)for(j=i;++j<l;)n[i*l+j]=h(s[i],s[j]);return java.util.Arrays.stream(n).max().getAsInt();


                          You can Try It Online!



                          Ungolfed and commented:



                          // Encode the character (swap case)
                          char e(boolean w, char c) 96 < c & c < 123) ? c ^ 32 : c);


                          // Pad the string to desired length
                          String p(String s, int l)
                          var p = "";
                          int m = s.length(), n = l / m, r = l % m, i = 0, j = 0;
                          var w = 1 < 0;
                          for (; i < n; ++i, w = !w)
                          for (char c : s.toCharArray())
                          p += e(w, c);
                          for (; j < r;)
                          p += e(w, s.charAt(j++));
                          return p;


                          // Calculate the actual hamming distance between two same-length strings
                          int d(String s, String t)
                          int l = s.length(), n = 0, i = 0;
                          for (; i < l;)
                          if (s.charAt(i) != t.charAt(i++))
                          ++n;
                          return n;

                          // Pad the strings as needed and return their hamming distance
                          int h(String s, String t)
                          int l = s.length(), m = t.length();
                          return l > m ? d(s, p(t, l)) : l < m ? d(p(s, m), t) : d(s, t);


                          // Dispatch the strings and gather their hamming distances, return the max
                          int h(String s)
                          int l = s.length, i = 0, j;
                          int n = new int[l * l];
                          for (; i < l; ++i)
                          for (j = i; ++j < l;)
                          n[i * l + j] = h(s[i], s[j]);
                          return java.util.Arrays.stream(n).max().getAsInt();



                          I know a better solution can be achieved, especially for the string pairing part.



                          EDIT: shave off 8 bytes by changing the size of the int array in hammingDistance() to the square of the numbe of strings given. It also fixes an ArrayIndexOutOfBounds thrown in one of the test cases.



                          EDIT 2: Saved 33 bytes thanks to Kevin Cruijssen's comments: class declaration removed, names shortened to 1 char, operators changed, etc.



                          EDIT 3: Save 1 byte and reach Satan-approved score by changing method with var-arg to array.



                          EDIT 4: Save another 50 bytes thanks to Kevin Cruijssen, again: update Java version from 8 to 10 to use var keyword, removed StringBuilder instance, etc.






                          share|improve this answer















                          Java 10, 748 740 667 666 616 bytes



                          This has to be the most dense and unreadable, yet the longest golf I ever came up with.



                          Call method h(String) with an explicit array (no var args): eg,



                          h(new String "a", "b", "c");


                          returns 1.



                          char e(boolean w,char c)96<c&c<123)?c^32:c);String p(String s,int l)var p="";int m=s.length(),n=l/m,r=l%m,i=0,j=0;var w=1<0;for(;i<n;++i,w=!w)for(char c:s.toCharArray())p+=e(w,c);for(;j<r;)p+=e(w,s.charAt(j++));return p;int d(String s,String t)int l=s.length(),n=0,i=0;for(;i<l;)if(s.charAt(i)!=t.charAt(i++))++n;return n;int h(String s,String t)int l=s.length(),m=t.length();return l>m?d(s,p(t,l)):l<m?d(p(s,m),t):d(s,t);int h(Strings)int l=s.length,i=0,j;intn=new int[l*l];for(;i<l;++i)for(j=i;++j<l;)n[i*l+j]=h(s[i],s[j]);return java.util.Arrays.stream(n).max().getAsInt();


                          You can Try It Online!



                          Ungolfed and commented:



                          // Encode the character (swap case)
                          char e(boolean w, char c) 96 < c & c < 123) ? c ^ 32 : c);


                          // Pad the string to desired length
                          String p(String s, int l)
                          var p = "";
                          int m = s.length(), n = l / m, r = l % m, i = 0, j = 0;
                          var w = 1 < 0;
                          for (; i < n; ++i, w = !w)
                          for (char c : s.toCharArray())
                          p += e(w, c);
                          for (; j < r;)
                          p += e(w, s.charAt(j++));
                          return p;


                          // Calculate the actual hamming distance between two same-length strings
                          int d(String s, String t)
                          int l = s.length(), n = 0, i = 0;
                          for (; i < l;)
                          if (s.charAt(i) != t.charAt(i++))
                          ++n;
                          return n;

                          // Pad the strings as needed and return their hamming distance
                          int h(String s, String t)
                          int l = s.length(), m = t.length();
                          return l > m ? d(s, p(t, l)) : l < m ? d(p(s, m), t) : d(s, t);


                          // Dispatch the strings and gather their hamming distances, return the max
                          int h(String s)
                          int l = s.length, i = 0, j;
                          int n = new int[l * l];
                          for (; i < l; ++i)
                          for (j = i; ++j < l;)
                          n[i * l + j] = h(s[i], s[j]);
                          return java.util.Arrays.stream(n).max().getAsInt();



                          I know a better solution can be achieved, especially for the string pairing part.



                          EDIT: shave off 8 bytes by changing the size of the int array in hammingDistance() to the square of the numbe of strings given. It also fixes an ArrayIndexOutOfBounds thrown in one of the test cases.



                          EDIT 2: Saved 33 bytes thanks to Kevin Cruijssen's comments: class declaration removed, names shortened to 1 char, operators changed, etc.



                          EDIT 3: Save 1 byte and reach Satan-approved score by changing method with var-arg to array.



                          EDIT 4: Save another 50 bytes thanks to Kevin Cruijssen, again: update Java version from 8 to 10 to use var keyword, removed StringBuilder instance, etc.







                          share|improve this answer














                          share|improve this answer



                          share|improve this answer








                          edited Aug 16 at 8:35

























                          answered Aug 14 at 9:39









                          joH1

                          286210




                          286210







                          • 1




                            I don't have a lot of time, but some basic things to golf: Drop the class, only the methods is enough. Change all method and variable names to single bytes.. So instead of hammingDistance use d or some other unused variable. Most of your && can be & and || can be |. c^' ' can be c^32. boolean w = false; can be boolean w=0>1;. i=0 in the loop initialization can be removed and change the ,i,j to ,i=0,j. ++j can be removed and ++ can be added to the .charAt(j++). .toString() can be +"". for(j=i+1;j<l;++j) can be for(j=0;++j<l;). Etc. etc.
                            – Kevin Cruijssen
                            Aug 14 at 13:38






                          • 1




                            Tips for golfing in Java and Tips for golfing in <all languages> might be interesting to read through as well. :)
                            – Kevin Cruijssen
                            Aug 14 at 13:41










                          • Thanks! That's some nice byte-lifting. Thanks for the links too, I'm taking a look at it and will edit asap!
                            – joH1
                            Aug 14 at 13:47






                          • 1




                            Upvoted for the Satan-approved score. xD Some more small things: StringBuilder can be StringBuffer (if you switch to Java 10 it could be var b=new StringBuffer(l);. The boolean and char can then also be var. If you don't have Java 10 locally, it is available on TIO). In addition, for(;i<n;++i)for(char c:s.toCharArray())b.append(e(w,c));w=!w; can be for(;i++<n;w=!w)for(char c:s.toCharArray())b.append(e(w,c));. And I'm pretty sure you can remove the StringBuffer completely and just use String and += instead of append.
                            – Kevin Cruijssen
                            Aug 14 at 15:13











                          • Man, some many months of clean code and good coding practices made me forget how to even golf! I'll update my answer and include TIO.
                            – joH1
                            Aug 16 at 7:15












                          • 1




                            I don't have a lot of time, but some basic things to golf: Drop the class, only the methods is enough. Change all method and variable names to single bytes.. So instead of hammingDistance use d or some other unused variable. Most of your && can be & and || can be |. c^' ' can be c^32. boolean w = false; can be boolean w=0>1;. i=0 in the loop initialization can be removed and change the ,i,j to ,i=0,j. ++j can be removed and ++ can be added to the .charAt(j++). .toString() can be +"". for(j=i+1;j<l;++j) can be for(j=0;++j<l;). Etc. etc.
                            – Kevin Cruijssen
                            Aug 14 at 13:38






                          • 1




                            Tips for golfing in Java and Tips for golfing in <all languages> might be interesting to read through as well. :)
                            – Kevin Cruijssen
                            Aug 14 at 13:41










                          • Thanks! That's some nice byte-lifting. Thanks for the links too, I'm taking a look at it and will edit asap!
                            – joH1
                            Aug 14 at 13:47






                          • 1




                            Upvoted for the Satan-approved score. xD Some more small things: StringBuilder can be StringBuffer (if you switch to Java 10 it could be var b=new StringBuffer(l);. The boolean and char can then also be var. If you don't have Java 10 locally, it is available on TIO). In addition, for(;i<n;++i)for(char c:s.toCharArray())b.append(e(w,c));w=!w; can be for(;i++<n;w=!w)for(char c:s.toCharArray())b.append(e(w,c));. And I'm pretty sure you can remove the StringBuffer completely and just use String and += instead of append.
                            – Kevin Cruijssen
                            Aug 14 at 15:13











                          • Man, some many months of clean code and good coding practices made me forget how to even golf! I'll update my answer and include TIO.
                            – joH1
                            Aug 16 at 7:15







                          1




                          1




                          I don't have a lot of time, but some basic things to golf: Drop the class, only the methods is enough. Change all method and variable names to single bytes.. So instead of hammingDistance use d or some other unused variable. Most of your && can be & and || can be |. c^' ' can be c^32. boolean w = false; can be boolean w=0>1;. i=0 in the loop initialization can be removed and change the ,i,j to ,i=0,j. ++j can be removed and ++ can be added to the .charAt(j++). .toString() can be +"". for(j=i+1;j<l;++j) can be for(j=0;++j<l;). Etc. etc.
                          – Kevin Cruijssen
                          Aug 14 at 13:38




                          I don't have a lot of time, but some basic things to golf: Drop the class, only the methods is enough. Change all method and variable names to single bytes.. So instead of hammingDistance use d or some other unused variable. Most of your && can be & and || can be |. c^' ' can be c^32. boolean w = false; can be boolean w=0>1;. i=0 in the loop initialization can be removed and change the ,i,j to ,i=0,j. ++j can be removed and ++ can be added to the .charAt(j++). .toString() can be +"". for(j=i+1;j<l;++j) can be for(j=0;++j<l;). Etc. etc.
                          – Kevin Cruijssen
                          Aug 14 at 13:38




                          1




                          1




                          Tips for golfing in Java and Tips for golfing in <all languages> might be interesting to read through as well. :)
                          – Kevin Cruijssen
                          Aug 14 at 13:41




                          Tips for golfing in Java and Tips for golfing in <all languages> might be interesting to read through as well. :)
                          – Kevin Cruijssen
                          Aug 14 at 13:41












                          Thanks! That's some nice byte-lifting. Thanks for the links too, I'm taking a look at it and will edit asap!
                          – joH1
                          Aug 14 at 13:47




                          Thanks! That's some nice byte-lifting. Thanks for the links too, I'm taking a look at it and will edit asap!
                          – joH1
                          Aug 14 at 13:47




                          1




                          1




                          Upvoted for the Satan-approved score. xD Some more small things: StringBuilder can be StringBuffer (if you switch to Java 10 it could be var b=new StringBuffer(l);. The boolean and char can then also be var. If you don't have Java 10 locally, it is available on TIO). In addition, for(;i<n;++i)for(char c:s.toCharArray())b.append(e(w,c));w=!w; can be for(;i++<n;w=!w)for(char c:s.toCharArray())b.append(e(w,c));. And I'm pretty sure you can remove the StringBuffer completely and just use String and += instead of append.
                          – Kevin Cruijssen
                          Aug 14 at 15:13





                          Upvoted for the Satan-approved score. xD Some more small things: StringBuilder can be StringBuffer (if you switch to Java 10 it could be var b=new StringBuffer(l);. The boolean and char can then also be var. If you don't have Java 10 locally, it is available on TIO). In addition, for(;i<n;++i)for(char c:s.toCharArray())b.append(e(w,c));w=!w; can be for(;i++<n;w=!w)for(char c:s.toCharArray())b.append(e(w,c));. And I'm pretty sure you can remove the StringBuffer completely and just use String and += instead of append.
                          – Kevin Cruijssen
                          Aug 14 at 15:13













                          Man, some many months of clean code and good coding practices made me forget how to even golf! I'll update my answer and include TIO.
                          – joH1
                          Aug 16 at 7:15




                          Man, some many months of clean code and good coding practices made me forget how to even golf! I'll update my answer and include TIO.
                          – joH1
                          Aug 16 at 7:15










                          up vote
                          1
                          down vote














                          05AB1E, 33 29 bytes



                          Ćü)€é©εćDš«s`g∍}®€¤‚ø€ζ€€Ë_Oà


                          Try it online or verify all test cases.



                          Can most likely be halved in byte-count, but it works..



                          Explanation:





                          Ć # Enclose the input-list (adding the first item to the end of the list)
                          # i.e. ['ABC1','abcD','abCd32e'] → ['ABC1','abcD','abCd32e','ABC1']
                          ü) # Pair-vectorize each of them
                          # i.e. ['ABC1','abcD','abCd32e','ABC1']
                          # → [['ABC1','abcD'],['abcD','abCd32e'],['abCd32e','ABC1']]
                          ێ # Sort each pair by length
                          # i.e. [['ABC1','abcD'],['abcD','abCd32e'],['abCd32e','ABC1']]
                          # → [['ABC1','abcD'],['abcD','abCd32e'],['ABC1','abCd32e']]
                          © # Store this list in the register to re-use later on
                          ε } # Map each inner list in this list to:
                          ć # Head extracted
                          # i.e. ['abcD','abCd32e'] → 'abcD' and ['abCd32e']
                          DÅ¡ # Duplicate it, and swap the capitalization of the copy
                          # i.e. 'abcD' → 'ABCd'
                          « # Then merge it together
                          # i.e. 'abcD' and 'ABCd' → 'abcDABCd'
                          s` # Swap so the tail-list is at the top of the stack, and get it's single item
                          # i.e. ['abCd32e'] → 'abCd32e'
                          g # Get the length of that
                          # i.e. 'abCd32e' → 7
                          ∍ # Extend/shorten the string to that length
                          # i.e. 'abcDABCd' and 7 → 'abcDABC'
                          ® # Get the saved list from the register again
                          €¤ # Get the tail from each
                          # i.e. [['ABC1','abcD'],['abcD','abCd32e'],['abCd32e','ABC1']]
                          # → ['abcD','abCd32e','abCd32e']
                          ‚ # Pair it with the other list
                          # i.e. ['ABC1','abcDABC','ABC1abc'] and ['abcD','abCd32e','abCd32e']
                          # → [['ABC1','abcDABC','ABC1abc'],['abcD','abCd32e','abCd32e']]
                          ø # Zip it, swapping rows / columns
                          # i.e. [['ABC1','abcDABC','ABC1abc'],['abcD','abCd32e','abCd32e']]
                          # → [['ABC1','abcD'],['abcDABC','abCd32e'],['ABC1abc','abCd32e']]
                          €ζ # And then zip each pair again
                          # i.e. [['ABC1','abcD'],['abcDABC','abCd32e'],['ABC1abc','abCd32e']]
                          # → [['Aa','Bb','Cc','1D'],['aa','bb','cC','Dd','A3','B2','Ce'],['Aa','Bb','CC','1d','a3','b2','ce']]
                          € # Then for each inner list
                          € # And for each inner string
                          Ë # Check if all characters are the same
                          # i.e. 'aa' → 1
                          # i.e. 'cC' → 0
                          _ # And inverse the booleans
                          # i.e. [['Aa','Bb','Cc','1D'],['aa','bb','cC','Dd','A3','B2','Ce'],['Aa','Bb','CC','1d','a3','b2','ce']]
                          # → [[1,1,1,1],[0,0,1,1,1,1,1],[1,1,0,1,1,1,1]]
                          O # Then sum each inner list
                          # i.e. [[1,1,1,1],[0,0,1,1,1,1,1],[1,1,0,1,1,1,1]] → [4,5,6]
                          à # And take the max as result
                          # i.e. [4,5,6] → 6





                          share|improve this answer
























                            up vote
                            1
                            down vote














                            05AB1E, 33 29 bytes



                            Ćü)€é©εćDš«s`g∍}®€¤‚ø€ζ€€Ë_Oà


                            Try it online or verify all test cases.



                            Can most likely be halved in byte-count, but it works..



                            Explanation:





                            Ć # Enclose the input-list (adding the first item to the end of the list)
                            # i.e. ['ABC1','abcD','abCd32e'] → ['ABC1','abcD','abCd32e','ABC1']
                            ü) # Pair-vectorize each of them
                            # i.e. ['ABC1','abcD','abCd32e','ABC1']
                            # → [['ABC1','abcD'],['abcD','abCd32e'],['abCd32e','ABC1']]
                            ێ # Sort each pair by length
                            # i.e. [['ABC1','abcD'],['abcD','abCd32e'],['abCd32e','ABC1']]
                            # → [['ABC1','abcD'],['abcD','abCd32e'],['ABC1','abCd32e']]
                            © # Store this list in the register to re-use later on
                            ε } # Map each inner list in this list to:
                            ć # Head extracted
                            # i.e. ['abcD','abCd32e'] → 'abcD' and ['abCd32e']
                            DÅ¡ # Duplicate it, and swap the capitalization of the copy
                            # i.e. 'abcD' → 'ABCd'
                            « # Then merge it together
                            # i.e. 'abcD' and 'ABCd' → 'abcDABCd'
                            s` # Swap so the tail-list is at the top of the stack, and get it's single item
                            # i.e. ['abCd32e'] → 'abCd32e'
                            g # Get the length of that
                            # i.e. 'abCd32e' → 7
                            ∍ # Extend/shorten the string to that length
                            # i.e. 'abcDABCd' and 7 → 'abcDABC'
                            ® # Get the saved list from the register again
                            €¤ # Get the tail from each
                            # i.e. [['ABC1','abcD'],['abcD','abCd32e'],['abCd32e','ABC1']]
                            # → ['abcD','abCd32e','abCd32e']
                            ‚ # Pair it with the other list
                            # i.e. ['ABC1','abcDABC','ABC1abc'] and ['abcD','abCd32e','abCd32e']
                            # → [['ABC1','abcDABC','ABC1abc'],['abcD','abCd32e','abCd32e']]
                            ø # Zip it, swapping rows / columns
                            # i.e. [['ABC1','abcDABC','ABC1abc'],['abcD','abCd32e','abCd32e']]
                            # → [['ABC1','abcD'],['abcDABC','abCd32e'],['ABC1abc','abCd32e']]
                            €ζ # And then zip each pair again
                            # i.e. [['ABC1','abcD'],['abcDABC','abCd32e'],['ABC1abc','abCd32e']]
                            # → [['Aa','Bb','Cc','1D'],['aa','bb','cC','Dd','A3','B2','Ce'],['Aa','Bb','CC','1d','a3','b2','ce']]
                            € # Then for each inner list
                            € # And for each inner string
                            Ë # Check if all characters are the same
                            # i.e. 'aa' → 1
                            # i.e. 'cC' → 0
                            _ # And inverse the booleans
                            # i.e. [['Aa','Bb','Cc','1D'],['aa','bb','cC','Dd','A3','B2','Ce'],['Aa','Bb','CC','1d','a3','b2','ce']]
                            # → [[1,1,1,1],[0,0,1,1,1,1,1],[1,1,0,1,1,1,1]]
                            O # Then sum each inner list
                            # i.e. [[1,1,1,1],[0,0,1,1,1,1,1],[1,1,0,1,1,1,1]] → [4,5,6]
                            à # And take the max as result
                            # i.e. [4,5,6] → 6





                            share|improve this answer






















                              up vote
                              1
                              down vote










                              up vote
                              1
                              down vote










                              05AB1E, 33 29 bytes



                              Ćü)€é©εćDš«s`g∍}®€¤‚ø€ζ€€Ë_Oà


                              Try it online or verify all test cases.



                              Can most likely be halved in byte-count, but it works..



                              Explanation:





                              Ć # Enclose the input-list (adding the first item to the end of the list)
                              # i.e. ['ABC1','abcD','abCd32e'] → ['ABC1','abcD','abCd32e','ABC1']
                              ü) # Pair-vectorize each of them
                              # i.e. ['ABC1','abcD','abCd32e','ABC1']
                              # → [['ABC1','abcD'],['abcD','abCd32e'],['abCd32e','ABC1']]
                              ێ # Sort each pair by length
                              # i.e. [['ABC1','abcD'],['abcD','abCd32e'],['abCd32e','ABC1']]
                              # → [['ABC1','abcD'],['abcD','abCd32e'],['ABC1','abCd32e']]
                              © # Store this list in the register to re-use later on
                              ε } # Map each inner list in this list to:
                              ć # Head extracted
                              # i.e. ['abcD','abCd32e'] → 'abcD' and ['abCd32e']
                              DÅ¡ # Duplicate it, and swap the capitalization of the copy
                              # i.e. 'abcD' → 'ABCd'
                              « # Then merge it together
                              # i.e. 'abcD' and 'ABCd' → 'abcDABCd'
                              s` # Swap so the tail-list is at the top of the stack, and get it's single item
                              # i.e. ['abCd32e'] → 'abCd32e'
                              g # Get the length of that
                              # i.e. 'abCd32e' → 7
                              ∍ # Extend/shorten the string to that length
                              # i.e. 'abcDABCd' and 7 → 'abcDABC'
                              ® # Get the saved list from the register again
                              €¤ # Get the tail from each
                              # i.e. [['ABC1','abcD'],['abcD','abCd32e'],['abCd32e','ABC1']]
                              # → ['abcD','abCd32e','abCd32e']
                              ‚ # Pair it with the other list
                              # i.e. ['ABC1','abcDABC','ABC1abc'] and ['abcD','abCd32e','abCd32e']
                              # → [['ABC1','abcDABC','ABC1abc'],['abcD','abCd32e','abCd32e']]
                              ø # Zip it, swapping rows / columns
                              # i.e. [['ABC1','abcDABC','ABC1abc'],['abcD','abCd32e','abCd32e']]
                              # → [['ABC1','abcD'],['abcDABC','abCd32e'],['ABC1abc','abCd32e']]
                              €ζ # And then zip each pair again
                              # i.e. [['ABC1','abcD'],['abcDABC','abCd32e'],['ABC1abc','abCd32e']]
                              # → [['Aa','Bb','Cc','1D'],['aa','bb','cC','Dd','A3','B2','Ce'],['Aa','Bb','CC','1d','a3','b2','ce']]
                              € # Then for each inner list
                              € # And for each inner string
                              Ë # Check if all characters are the same
                              # i.e. 'aa' → 1
                              # i.e. 'cC' → 0
                              _ # And inverse the booleans
                              # i.e. [['Aa','Bb','Cc','1D'],['aa','bb','cC','Dd','A3','B2','Ce'],['Aa','Bb','CC','1d','a3','b2','ce']]
                              # → [[1,1,1,1],[0,0,1,1,1,1,1],[1,1,0,1,1,1,1]]
                              O # Then sum each inner list
                              # i.e. [[1,1,1,1],[0,0,1,1,1,1,1],[1,1,0,1,1,1,1]] → [4,5,6]
                              à # And take the max as result
                              # i.e. [4,5,6] → 6





                              share|improve this answer













                              05AB1E, 33 29 bytes



                              Ćü)€é©εćDš«s`g∍}®€¤‚ø€ζ€€Ë_Oà


                              Try it online or verify all test cases.



                              Can most likely be halved in byte-count, but it works..



                              Explanation:





                              Ć # Enclose the input-list (adding the first item to the end of the list)
                              # i.e. ['ABC1','abcD','abCd32e'] → ['ABC1','abcD','abCd32e','ABC1']
                              ü) # Pair-vectorize each of them
                              # i.e. ['ABC1','abcD','abCd32e','ABC1']
                              # → [['ABC1','abcD'],['abcD','abCd32e'],['abCd32e','ABC1']]
                              ێ # Sort each pair by length
                              # i.e. [['ABC1','abcD'],['abcD','abCd32e'],['abCd32e','ABC1']]
                              # → [['ABC1','abcD'],['abcD','abCd32e'],['ABC1','abCd32e']]
                              © # Store this list in the register to re-use later on
                              ε } # Map each inner list in this list to:
                              ć # Head extracted
                              # i.e. ['abcD','abCd32e'] → 'abcD' and ['abCd32e']
                              DÅ¡ # Duplicate it, and swap the capitalization of the copy
                              # i.e. 'abcD' → 'ABCd'
                              « # Then merge it together
                              # i.e. 'abcD' and 'ABCd' → 'abcDABCd'
                              s` # Swap so the tail-list is at the top of the stack, and get it's single item
                              # i.e. ['abCd32e'] → 'abCd32e'
                              g # Get the length of that
                              # i.e. 'abCd32e' → 7
                              ∍ # Extend/shorten the string to that length
                              # i.e. 'abcDABCd' and 7 → 'abcDABC'
                              ® # Get the saved list from the register again
                              €¤ # Get the tail from each
                              # i.e. [['ABC1','abcD'],['abcD','abCd32e'],['abCd32e','ABC1']]
                              # → ['abcD','abCd32e','abCd32e']
                              ‚ # Pair it with the other list
                              # i.e. ['ABC1','abcDABC','ABC1abc'] and ['abcD','abCd32e','abCd32e']
                              # → [['ABC1','abcDABC','ABC1abc'],['abcD','abCd32e','abCd32e']]
                              ø # Zip it, swapping rows / columns
                              # i.e. [['ABC1','abcDABC','ABC1abc'],['abcD','abCd32e','abCd32e']]
                              # → [['ABC1','abcD'],['abcDABC','abCd32e'],['ABC1abc','abCd32e']]
                              €ζ # And then zip each pair again
                              # i.e. [['ABC1','abcD'],['abcDABC','abCd32e'],['ABC1abc','abCd32e']]
                              # → [['Aa','Bb','Cc','1D'],['aa','bb','cC','Dd','A3','B2','Ce'],['Aa','Bb','CC','1d','a3','b2','ce']]
                              € # Then for each inner list
                              € # And for each inner string
                              Ë # Check if all characters are the same
                              # i.e. 'aa' → 1
                              # i.e. 'cC' → 0
                              _ # And inverse the booleans
                              # i.e. [['Aa','Bb','Cc','1D'],['aa','bb','cC','Dd','A3','B2','Ce'],['Aa','Bb','CC','1d','a3','b2','ce']]
                              # → [[1,1,1,1],[0,0,1,1,1,1,1],[1,1,0,1,1,1,1]]
                              O # Then sum each inner list
                              # i.e. [[1,1,1,1],[0,0,1,1,1,1,1],[1,1,0,1,1,1,1]] → [4,5,6]
                              à # And take the max as result
                              # i.e. [4,5,6] → 6






                              share|improve this answer












                              share|improve this answer



                              share|improve this answer










                              answered Aug 14 at 9:30









                              Kevin Cruijssen

                              29.4k548162




                              29.4k548162




















                                  up vote
                                  1
                                  down vote













                                  Java 11, 387 bytes





                                  a->int l=a.length,i=l,t,j=0,C=new int[l];var p=new String[l][2];for(;i-->0;p[i][0]=a[t>0?i:j],p[i][1]=a[t>0?j:i])t=a[i].length()<a[j=-~i%l].length()?1:0;i=0;for(var P:p)var s="";for(var x:P[0].getBytes())s+=(char)(x>64&x<91for(int c:C)j=c>j?c:j;return j;


                                  Try it online. (NOTE: Since Java 11 isn't on TIO yet, String.repeat(int) has been emulated as repeat(String,int) for the same byte-count.)



                                  Explanation:



                                  a-> // Method with String-array parameter and integer return-type
                                  int l=a.length, // Length of the input-array
                                  i=l, // Index-integer, starting at the length
                                  t,j=0, // Temp-integers
                                  C=new int[l]; // Count-array the same size as the input
                                  var p=new String[l][2]; // String-pairs array the same size as the input
                                  for(;i-->0 // Loop `i` in the range [`l`, 0)
                                  ; // After every iteration:
                                  p[i][0]= // Set the first String of the pair at index `i` to:
                                  a[t>0?i:j],// The smallest of the `i`'th or `j`'th Strings of the input-array
                                  p[i][1]= // And set the second String of the pair at index `i` to:
                                  a[t>0?j:i])// The largest of the `i`'th or `j`'th Strings of the input-array
                                  t=a[i].length()< // If the length of the `i`'th item is smaller than
                                  a[j=-~i%l].length()?// the length of the `i+1`'th item
                                  // (and set `j` to this `i+1` with wrap-around to 0 for the last item
                                  1 // Set `t` to 1 as flag
                                  : // Else:
                                  0; // Set `t` to 0 as flag
                                  // We've now created the String pairs, where each pair is sorted by length
                                  i=0; // Reset `i` to 0
                                  for(var P:p) // Loop over the pairs
                                  var s=""; // Temp-String starting empty
                                  for(var x:P[0].getBytes())
                                  // Loop over the characters of the first String of the pair
                                  s+= // Append the temp-String with:
                                  (char)(x>64&x<91 // After every iteration of the outer loop,
                                  // increase `i` by 1 for the next iteration
                                  for(int c:C) // Now loop over the calculated counts
                                  j=c>j?c:j; // And set `j` to the maximum
                                  return j; // And finally return this maximum `j` as result





                                  share|improve this answer
























                                    up vote
                                    1
                                    down vote













                                    Java 11, 387 bytes





                                    a->int l=a.length,i=l,t,j=0,C=new int[l];var p=new String[l][2];for(;i-->0;p[i][0]=a[t>0?i:j],p[i][1]=a[t>0?j:i])t=a[i].length()<a[j=-~i%l].length()?1:0;i=0;for(var P:p)var s="";for(var x:P[0].getBytes())s+=(char)(x>64&x<91for(int c:C)j=c>j?c:j;return j;


                                    Try it online. (NOTE: Since Java 11 isn't on TIO yet, String.repeat(int) has been emulated as repeat(String,int) for the same byte-count.)



                                    Explanation:



                                    a-> // Method with String-array parameter and integer return-type
                                    int l=a.length, // Length of the input-array
                                    i=l, // Index-integer, starting at the length
                                    t,j=0, // Temp-integers
                                    C=new int[l]; // Count-array the same size as the input
                                    var p=new String[l][2]; // String-pairs array the same size as the input
                                    for(;i-->0 // Loop `i` in the range [`l`, 0)
                                    ; // After every iteration:
                                    p[i][0]= // Set the first String of the pair at index `i` to:
                                    a[t>0?i:j],// The smallest of the `i`'th or `j`'th Strings of the input-array
                                    p[i][1]= // And set the second String of the pair at index `i` to:
                                    a[t>0?j:i])// The largest of the `i`'th or `j`'th Strings of the input-array
                                    t=a[i].length()< // If the length of the `i`'th item is smaller than
                                    a[j=-~i%l].length()?// the length of the `i+1`'th item
                                    // (and set `j` to this `i+1` with wrap-around to 0 for the last item
                                    1 // Set `t` to 1 as flag
                                    : // Else:
                                    0; // Set `t` to 0 as flag
                                    // We've now created the String pairs, where each pair is sorted by length
                                    i=0; // Reset `i` to 0
                                    for(var P:p) // Loop over the pairs
                                    var s=""; // Temp-String starting empty
                                    for(var x:P[0].getBytes())
                                    // Loop over the characters of the first String of the pair
                                    s+= // Append the temp-String with:
                                    (char)(x>64&x<91 // After every iteration of the outer loop,
                                    // increase `i` by 1 for the next iteration
                                    for(int c:C) // Now loop over the calculated counts
                                    j=c>j?c:j; // And set `j` to the maximum
                                    return j; // And finally return this maximum `j` as result





                                    share|improve this answer






















                                      up vote
                                      1
                                      down vote










                                      up vote
                                      1
                                      down vote









                                      Java 11, 387 bytes





                                      a->int l=a.length,i=l,t,j=0,C=new int[l];var p=new String[l][2];for(;i-->0;p[i][0]=a[t>0?i:j],p[i][1]=a[t>0?j:i])t=a[i].length()<a[j=-~i%l].length()?1:0;i=0;for(var P:p)var s="";for(var x:P[0].getBytes())s+=(char)(x>64&x<91for(int c:C)j=c>j?c:j;return j;


                                      Try it online. (NOTE: Since Java 11 isn't on TIO yet, String.repeat(int) has been emulated as repeat(String,int) for the same byte-count.)



                                      Explanation:



                                      a-> // Method with String-array parameter and integer return-type
                                      int l=a.length, // Length of the input-array
                                      i=l, // Index-integer, starting at the length
                                      t,j=0, // Temp-integers
                                      C=new int[l]; // Count-array the same size as the input
                                      var p=new String[l][2]; // String-pairs array the same size as the input
                                      for(;i-->0 // Loop `i` in the range [`l`, 0)
                                      ; // After every iteration:
                                      p[i][0]= // Set the first String of the pair at index `i` to:
                                      a[t>0?i:j],// The smallest of the `i`'th or `j`'th Strings of the input-array
                                      p[i][1]= // And set the second String of the pair at index `i` to:
                                      a[t>0?j:i])// The largest of the `i`'th or `j`'th Strings of the input-array
                                      t=a[i].length()< // If the length of the `i`'th item is smaller than
                                      a[j=-~i%l].length()?// the length of the `i+1`'th item
                                      // (and set `j` to this `i+1` with wrap-around to 0 for the last item
                                      1 // Set `t` to 1 as flag
                                      : // Else:
                                      0; // Set `t` to 0 as flag
                                      // We've now created the String pairs, where each pair is sorted by length
                                      i=0; // Reset `i` to 0
                                      for(var P:p) // Loop over the pairs
                                      var s=""; // Temp-String starting empty
                                      for(var x:P[0].getBytes())
                                      // Loop over the characters of the first String of the pair
                                      s+= // Append the temp-String with:
                                      (char)(x>64&x<91 // After every iteration of the outer loop,
                                      // increase `i` by 1 for the next iteration
                                      for(int c:C) // Now loop over the calculated counts
                                      j=c>j?c:j; // And set `j` to the maximum
                                      return j; // And finally return this maximum `j` as result





                                      share|improve this answer












                                      Java 11, 387 bytes





                                      a->int l=a.length,i=l,t,j=0,C=new int[l];var p=new String[l][2];for(;i-->0;p[i][0]=a[t>0?i:j],p[i][1]=a[t>0?j:i])t=a[i].length()<a[j=-~i%l].length()?1:0;i=0;for(var P:p)var s="";for(var x:P[0].getBytes())s+=(char)(x>64&x<91for(int c:C)j=c>j?c:j;return j;


                                      Try it online. (NOTE: Since Java 11 isn't on TIO yet, String.repeat(int) has been emulated as repeat(String,int) for the same byte-count.)



                                      Explanation:



                                      a-> // Method with String-array parameter and integer return-type
                                      int l=a.length, // Length of the input-array
                                      i=l, // Index-integer, starting at the length
                                      t,j=0, // Temp-integers
                                      C=new int[l]; // Count-array the same size as the input
                                      var p=new String[l][2]; // String-pairs array the same size as the input
                                      for(;i-->0 // Loop `i` in the range [`l`, 0)
                                      ; // After every iteration:
                                      p[i][0]= // Set the first String of the pair at index `i` to:
                                      a[t>0?i:j],// The smallest of the `i`'th or `j`'th Strings of the input-array
                                      p[i][1]= // And set the second String of the pair at index `i` to:
                                      a[t>0?j:i])// The largest of the `i`'th or `j`'th Strings of the input-array
                                      t=a[i].length()< // If the length of the `i`'th item is smaller than
                                      a[j=-~i%l].length()?// the length of the `i+1`'th item
                                      // (and set `j` to this `i+1` with wrap-around to 0 for the last item
                                      1 // Set `t` to 1 as flag
                                      : // Else:
                                      0; // Set `t` to 0 as flag
                                      // We've now created the String pairs, where each pair is sorted by length
                                      i=0; // Reset `i` to 0
                                      for(var P:p) // Loop over the pairs
                                      var s=""; // Temp-String starting empty
                                      for(var x:P[0].getBytes())
                                      // Loop over the characters of the first String of the pair
                                      s+= // Append the temp-String with:
                                      (char)(x>64&x<91 // After every iteration of the outer loop,
                                      // increase `i` by 1 for the next iteration
                                      for(int c:C) // Now loop over the calculated counts
                                      j=c>j?c:j; // And set `j` to the maximum
                                      return j; // And finally return this maximum `j` as result






                                      share|improve this answer












                                      share|improve this answer



                                      share|improve this answer










                                      answered Aug 14 at 14:40









                                      Kevin Cruijssen

                                      29.4k548162




                                      29.4k548162




















                                          up vote
                                          1
                                          down vote














                                          R, 173 bytes





                                          function(x,U=utf8ToInt,N=nchar)max(combn(x,2,function(z,v=z[order(N(z))])sum(U(substr(Reduce(paste0,rep(c(v[1],chartr('A-Za-z','a-zA-Z',v[1])),n<-N(v[2]))),1,n))!=U(v[2]))))


                                          Try it online!



                                          @ngm : I tried my best to golf your code (with my heavy customizations of course) but, as you well know, R is not very golfy in manipulating strings :P






                                          share|improve this answer






















                                          • I bet this can be sub 150 bytes, but I'm not sure how just yet.
                                            – Giuseppe
                                            Aug 15 at 17:50










                                          • @Giuseppe: I suspect that too... but I'm not really good in writing short strings manipulation codes and R doesn't help me very much either :D
                                            – digEmAll
                                            Aug 15 at 18:56










                                          • @digEmAll I'm not going to try to solve my own challenge, but few possibilities might include outer to get all the combinations, and doing modular arithmetic on the code points in lieu of chartr.
                                            – ngm
                                            Aug 16 at 0:33










                                          • @ngm: possible...I discarded the arithmetic approach because I couldn't find a short solution/formula to change the case for letters without touching the numbers...
                                            – digEmAll
                                            Aug 16 at 6:18














                                          up vote
                                          1
                                          down vote














                                          R, 173 bytes





                                          function(x,U=utf8ToInt,N=nchar)max(combn(x,2,function(z,v=z[order(N(z))])sum(U(substr(Reduce(paste0,rep(c(v[1],chartr('A-Za-z','a-zA-Z',v[1])),n<-N(v[2]))),1,n))!=U(v[2]))))


                                          Try it online!



                                          @ngm : I tried my best to golf your code (with my heavy customizations of course) but, as you well know, R is not very golfy in manipulating strings :P






                                          share|improve this answer






















                                          • I bet this can be sub 150 bytes, but I'm not sure how just yet.
                                            – Giuseppe
                                            Aug 15 at 17:50










                                          • @Giuseppe: I suspect that too... but I'm not really good in writing short strings manipulation codes and R doesn't help me very much either :D
                                            – digEmAll
                                            Aug 15 at 18:56










                                          • @digEmAll I'm not going to try to solve my own challenge, but few possibilities might include outer to get all the combinations, and doing modular arithmetic on the code points in lieu of chartr.
                                            – ngm
                                            Aug 16 at 0:33










                                          • @ngm: possible...I discarded the arithmetic approach because I couldn't find a short solution/formula to change the case for letters without touching the numbers...
                                            – digEmAll
                                            Aug 16 at 6:18












                                          up vote
                                          1
                                          down vote










                                          up vote
                                          1
                                          down vote










                                          R, 173 bytes





                                          function(x,U=utf8ToInt,N=nchar)max(combn(x,2,function(z,v=z[order(N(z))])sum(U(substr(Reduce(paste0,rep(c(v[1],chartr('A-Za-z','a-zA-Z',v[1])),n<-N(v[2]))),1,n))!=U(v[2]))))


                                          Try it online!



                                          @ngm : I tried my best to golf your code (with my heavy customizations of course) but, as you well know, R is not very golfy in manipulating strings :P






                                          share|improve this answer















                                          R, 173 bytes





                                          function(x,U=utf8ToInt,N=nchar)max(combn(x,2,function(z,v=z[order(N(z))])sum(U(substr(Reduce(paste0,rep(c(v[1],chartr('A-Za-z','a-zA-Z',v[1])),n<-N(v[2]))),1,n))!=U(v[2]))))


                                          Try it online!



                                          @ngm : I tried my best to golf your code (with my heavy customizations of course) but, as you well know, R is not very golfy in manipulating strings :P







                                          share|improve this answer














                                          share|improve this answer



                                          share|improve this answer








                                          edited Aug 15 at 17:06

























                                          answered Aug 15 at 7:41









                                          digEmAll

                                          1,71147




                                          1,71147











                                          • I bet this can be sub 150 bytes, but I'm not sure how just yet.
                                            – Giuseppe
                                            Aug 15 at 17:50










                                          • @Giuseppe: I suspect that too... but I'm not really good in writing short strings manipulation codes and R doesn't help me very much either :D
                                            – digEmAll
                                            Aug 15 at 18:56










                                          • @digEmAll I'm not going to try to solve my own challenge, but few possibilities might include outer to get all the combinations, and doing modular arithmetic on the code points in lieu of chartr.
                                            – ngm
                                            Aug 16 at 0:33










                                          • @ngm: possible...I discarded the arithmetic approach because I couldn't find a short solution/formula to change the case for letters without touching the numbers...
                                            – digEmAll
                                            Aug 16 at 6:18
















                                          • I bet this can be sub 150 bytes, but I'm not sure how just yet.
                                            – Giuseppe
                                            Aug 15 at 17:50










                                          • @Giuseppe: I suspect that too... but I'm not really good in writing short strings manipulation codes and R doesn't help me very much either :D
                                            – digEmAll
                                            Aug 15 at 18:56










                                          • @digEmAll I'm not going to try to solve my own challenge, but few possibilities might include outer to get all the combinations, and doing modular arithmetic on the code points in lieu of chartr.
                                            – ngm
                                            Aug 16 at 0:33










                                          • @ngm: possible...I discarded the arithmetic approach because I couldn't find a short solution/formula to change the case for letters without touching the numbers...
                                            – digEmAll
                                            Aug 16 at 6:18















                                          I bet this can be sub 150 bytes, but I'm not sure how just yet.
                                          – Giuseppe
                                          Aug 15 at 17:50




                                          I bet this can be sub 150 bytes, but I'm not sure how just yet.
                                          – Giuseppe
                                          Aug 15 at 17:50












                                          @Giuseppe: I suspect that too... but I'm not really good in writing short strings manipulation codes and R doesn't help me very much either :D
                                          – digEmAll
                                          Aug 15 at 18:56




                                          @Giuseppe: I suspect that too... but I'm not really good in writing short strings manipulation codes and R doesn't help me very much either :D
                                          – digEmAll
                                          Aug 15 at 18:56












                                          @digEmAll I'm not going to try to solve my own challenge, but few possibilities might include outer to get all the combinations, and doing modular arithmetic on the code points in lieu of chartr.
                                          – ngm
                                          Aug 16 at 0:33




                                          @digEmAll I'm not going to try to solve my own challenge, but few possibilities might include outer to get all the combinations, and doing modular arithmetic on the code points in lieu of chartr.
                                          – ngm
                                          Aug 16 at 0:33












                                          @ngm: possible...I discarded the arithmetic approach because I couldn't find a short solution/formula to change the case for letters without touching the numbers...
                                          – digEmAll
                                          Aug 16 at 6:18




                                          @ngm: possible...I discarded the arithmetic approach because I couldn't find a short solution/formula to change the case for letters without touching the numbers...
                                          – digEmAll
                                          Aug 16 at 6:18

















                                           

                                          draft saved


                                          draft discarded















































                                           


                                          draft saved


                                          draft discarded














                                          StackExchange.ready(
                                          function ()
                                          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcodegolf.stackexchange.com%2fquestions%2f170581%2fmaximum-hamming-distance-among-a-list-of-padded-strings%23new-answer', 'question_page');

                                          );

                                          Post as a guest













































































                                          Comments

                                          Popular posts from this blog

                                          What does second last employer means? [closed]

                                          Installing NextGIS Connect into QGIS 3?

                                          One-line joke