Mixing
Make an alphabeTrie
Clash Royale CLAN TAG#URR8PPP
up vote
29
down vote
favorite
Consider the following alphabetically sorted list of words:
balderdash
ballet
balloonfish
balloonist
ballot
brooding
broom
All of the words start with b, and the first 5 start with bal. If we just look at the first 2 words:
balderdash
ballet
we could write instead:
balderdash
+let
where the ' ' is used where a word shares a prefix character with the previous word; except for the '+' character which indicates the LAST character where the second word shares a prefix with the previous word.
This is a sort of 'trie' visualization: the parent is 'bal', and has 2 descendants: 'derdash' and 'let'.
With a longer list, such as:
balderdash
ballet
brooding
we can additionally use the pipe character '|' to make it clearer where the shared prefix ends, as follows:
balderdash
| +let
+rooding
and the equivalent tree would have a root of 'b' having two children: the subtree having root 'al' and and its two children 'derdash' and 'let'; and 'rooding'.
If we apply this strategy to our original list,
balderdash
ballet
balloonfish
balloonist
ballot
brooding
broom
we get output that looks like:
balderdash
| +let
| +oonfish
| | +ist
| +t
+rooding
+m
If two consecutive words in the list have no shared prefix, no special characters are substituted; e.g. for the list:
broom
brood
crude
crumb
we want the output:
broom
+d
crude
+mb
Input
The words in the input will be made up of only alphanumerics (no spaces or punctuation); this may be in the form of a list of strings, a single string, or any other reasonable approach, as long as you specify your chosen format. No two consecutive words will be the same. The list will be alphabetically sorted.
Output
Your output can contain trailing whitespace per line or in total, but no leading whitespace. A list of strings or similar would also be acceptable.
This is code-golf; the shortest code in each language retains bragging rights. The usual prohibitions against loopholes apply.
Test Cases
Input:
apogee
apology
app
apple
applique
apply
apt
Output:
apogee
|+logy
+p
|+le
| +ique
| +y
+t
Input:
balderdash
ballet
balloonfish
balloonist
ballot
brooding
broom
donald
donatella
donna
dont
dumb
Output:
balderdash
| +let
| +oonfish
| | +ist
| +t
+rooding
+m
donald
| |+tella
| +na
| +t
+umb
code-golf string
edited Aug 13 at 19:30
Mr. Xcoder
30.1k757193
asked Aug 13 at 19:12
Chas Brown
4,1361319
 |Â
show 3 more comments
up vote
29
down vote
favorite
Consider the following alphabetically sorted list of words:
balderdash
ballet
balloonfish
balloonist
ballot
brooding
broom
All of the words start with b, and the first 5 start with bal. If we just look at the first 2 words:
balderdash
ballet
we could write instead:
balderdash
+let
where the ' ' is used where a word shares a prefix character with the previous word; except for the '+' character which indicates the LAST character where the second word shares a prefix with the previous word.
This is a sort of 'trie' visualization: the parent is 'bal', and has 2 descendants: 'derdash' and 'let'.
With a longer list, such as:
balderdash
ballet
brooding
we can additionally use the pipe character '|' to make it clearer where the shared prefix ends, as follows:
balderdash
| +let
+rooding
and the equivalent tree would have a root of 'b' having two children: the subtree having root 'al' and and its two children 'derdash' and 'let'; and 'rooding'.
If we apply this strategy to our original list,
balderdash
ballet
balloonfish
balloonist
ballot
brooding
broom
we get output that looks like:
balderdash
| +let
| +oonfish
| | +ist
| +t
+rooding
+m
If two consecutive words in the list have no shared prefix, no special characters are substituted; e.g. for the list:
broom
brood
crude
crumb
we want the output:
broom
+d
crude
+mb
Input
The words in the input will be made up of only alphanumerics (no spaces or punctuation); this may be in the form of a list of strings, a single string, or any other reasonable approach, as long as you specify your chosen format. No two consecutive words will be the same. The list will be alphabetically sorted.
Output
Your output can contain trailing whitespace per line or in total, but no leading whitespace. A list of strings or similar would also be acceptable.
This is code-golf; the shortest code in each language retains bragging rights. The usual prohibitions against loopholes apply.
Test Cases
Input:
apogee
apology
app
apple
applique
apply
apt
Output:
apogee
|+logy
+p
|+le
| +ique
| +y
+t
Input:
balderdash
ballet
balloonfish
balloonist
ballot
brooding
broom
donald
donatella
donna
dont
dumb
Output:
balderdash
| +let
| +oonfish
| | +ist
| +t
+rooding
+m
donald
| |+tella
| +na
| +t
+umb
code-golf string
edited Aug 13 at 19:30
Mr. Xcoder
30.1k757193
asked Aug 13 at 19:12
Chas Brown
4,1361319
What about the case where I have the wordballafterballoon. What output should we expect?
– Rushabh Mehta
Aug 13 at 20:05
@RushabhMehta I'm guessing you would just have a+under the firsto, but I didn't write the challenge so I'm not certain.
– Theo
Aug 13 at 20:12
5
@RushabhMehta The words are alphabetically sorted, so this won't happen.
– Neil
Aug 13 at 20:14
@Neil Oh good point
– Rushabh Mehta
Aug 13 at 20:18
2
The words in the input will be made up of only alphanumerics: does that really include digits, or did you mean alphabetic?
– Arnauld
Aug 14 at 1:40
 |Â
show 3 more comments
up vote
29
down vote
favorite
up vote
29
down vote
favorite
Consider the following alphabetically sorted list of words:
balderdash
ballet
balloonfish
balloonist
ballot
brooding
broom
All of the words start with b, and the first 5 start with bal. If we just look at the first 2 words:
balderdash
ballet
we could write instead:
balderdash
+let
where the ' ' is used where a word shares a prefix character with the previous word; except for the '+' character which indicates the LAST character where the second word shares a prefix with the previous word.
This is a sort of 'trie' visualization: the parent is 'bal', and has 2 descendants: 'derdash' and 'let'.
With a longer list, such as:
balderdash
ballet
brooding
we can additionally use the pipe character '|' to make it clearer where the shared prefix ends, as follows:
balderdash
| +let
+rooding
and the equivalent tree would have a root of 'b' having two children: the subtree having root 'al' and and its two children 'derdash' and 'let'; and 'rooding'.
If we apply this strategy to our original list,
balderdash
ballet
balloonfish
balloonist
ballot
brooding
broom
we get output that looks like:
balderdash
| +let
| +oonfish
| | +ist
| +t
+rooding
+m
If two consecutive words in the list have no shared prefix, no special characters are substituted; e.g. for the list:
broom
brood
crude
crumb
we want the output:
broom
+d
crude
+mb
Input
The words in the input will be made up of only alphanumerics (no spaces or punctuation); this may be in the form of a list of strings, a single string, or any other reasonable approach, as long as you specify your chosen format. No two consecutive words will be the same. The list will be alphabetically sorted.
Output
Your output can contain trailing whitespace per line or in total, but no leading whitespace. A list of strings or similar would also be acceptable.
This is code-golf; the shortest code in each language retains bragging rights. The usual prohibitions against loopholes apply.
Test Cases
Input:
apogee
apology
app
apple
applique
apply
apt
Output:
apogee
|+logy
+p
|+le
| +ique
| +y
+t
Input:
balderdash
ballet
balloonfish
balloonist
ballot
brooding
broom
donald
donatella
donna
dont
dumb
Output:
balderdash
| +let
| +oonfish
| | +ist
| +t
+rooding
+m
donald
| |+tella
| +na
| +t
+umb
code-golf string
edited Aug 13 at 19:30
Mr. Xcoder
30.1k757193
asked Aug 13 at 19:12
Chas Brown
4,1361319
Consider the following alphabetically sorted list of words:
balderdash
ballet
balloonfish
balloonist
ballot
brooding
broom
All of the words start with b, and the first 5 start with bal. If we just look at the first 2 words:
balderdash
ballet
we could write instead:
balderdash
+let
where the ' ' is used where a word shares a prefix character with the previous word; except for the '+' character which indicates the LAST character where the second word shares a prefix with the previous word.
This is a sort of 'trie' visualization: the parent is 'bal', and has 2 descendants: 'derdash' and 'let'.
With a longer list, such as:
balderdash
ballet
brooding
we can additionally use the pipe character '|' to make it clearer where the shared prefix ends, as follows:
balderdash
| +let
+rooding
and the equivalent tree would have a root of 'b' having two children: the subtree having root 'al' and and its two children 'derdash' and 'let'; and 'rooding'.
If we apply this strategy to our original list,
balderdash
ballet
balloonfish
balloonist
ballot
brooding
broom
we get output that looks like:
balderdash
| +let
| +oonfish
| | +ist
| +t
+rooding
+m
If two consecutive words in the list have no shared prefix, no special characters are substituted; e.g. for the list:
broom
brood
crude
crumb
we want the output:
broom
+d
crude
+mb
Input
The words in the input will be made up of only alphanumerics (no spaces or punctuation); this may be in the form of a list of strings, a single string, or any other reasonable approach, as long as you specify your chosen format. No two consecutive words will be the same. The list will be alphabetically sorted.
Output
Your output can contain trailing whitespace per line or in total, but no leading whitespace. A list of strings or similar would also be acceptable.
This is code-golf; the shortest code in each language retains bragging rights. The usual prohibitions against loopholes apply.
Test Cases
Input:
apogee
apology
app
apple
applique
apply
apt
Output:
apogee
|+logy
+p
|+le
| +ique
| +y
+t
Input:
balderdash
ballet
balloonfish
balloonist
ballot
brooding
broom
donald
donatella
donna
dont
dumb
Output:
balderdash
| +let
| +oonfish
| | +ist
| +t
+rooding
+m
donald
| |+tella
| +na
| +t
+umb
code-golf string
edited Aug 13 at 19:30
Mr. Xcoder
30.1k757193
asked Aug 13 at 19:12
Chas Brown
4,1361319
edited Aug 13 at 19:30
Mr. Xcoder
30.1k757193
edited Aug 13 at 19:30
Mr. Xcoder
30.1k757193
edited Aug 13 at 19:30
Mr. Xcoder
30.1k757193
30.1k757193
asked Aug 13 at 19:12
Chas Brown
4,1361319
asked Aug 13 at 19:12
Chas Brown
4,1361319
asked Aug 13 at 19:12
Chas Brown
4,1361319
4,1361319
What about the case where I have the wordballafterballoon. What output should we expect?
– Rushabh Mehta
Aug 13 at 20:05
@RushabhMehta I'm guessing you would just have a+under the firsto, but I didn't write the challenge so I'm not certain.
– Theo
Aug 13 at 20:12
5
@RushabhMehta The words are alphabetically sorted, so this won't happen.
– Neil
Aug 13 at 20:14
@Neil Oh good point
– Rushabh Mehta
Aug 13 at 20:18
2
The words in the input will be made up of only alphanumerics: does that really include digits, or did you mean alphabetic?
– Arnauld
Aug 14 at 1:40
 |Â
show 3 more comments
What about the case where I have the wordballafterballoon. What output should we expect?
– Rushabh Mehta
Aug 13 at 20:05
@RushabhMehta I'm guessing you would just have a+under the firsto, but I didn't write the challenge so I'm not certain.
– Theo
Aug 13 at 20:12
5
@RushabhMehta The words are alphabetically sorted, so this won't happen.
– Neil
Aug 13 at 20:14
@Neil Oh good point
– Rushabh Mehta
Aug 13 at 20:18
2
The words in the input will be made up of only alphanumerics: does that really include digits, or did you mean alphabetic?
– Arnauld
Aug 14 at 1:40
What about the case where I have the word
ball after balloon. What output should we expect?– Rushabh Mehta
Aug 13 at 20:05
What about the case where I have the word
ball after balloon. What output should we expect?– Rushabh Mehta
Aug 13 at 20:05
@RushabhMehta I'm guessing you would just have a
+ under the first o, but I didn't write the challenge so I'm not certain.– Theo
Aug 13 at 20:12
@RushabhMehta I'm guessing you would just have a
+ under the first o, but I didn't write the challenge so I'm not certain.– Theo
Aug 13 at 20:12
5
5
@RushabhMehta The words are alphabetically sorted, so this won't happen.
– Neil
Aug 13 at 20:14
@RushabhMehta The words are alphabetically sorted, so this won't happen.
– Neil
Aug 13 at 20:14
@Neil Oh good point
– Rushabh Mehta
Aug 13 at 20:18
@Neil Oh good point
– Rushabh Mehta
Aug 13 at 20:18
2
2
The words in the input will be made up of only alphanumerics: does that really include digits, or did you mean alphabetic?
– Arnauld
Aug 14 at 1:40
The words in the input will be made up of only alphanumerics: does that really include digits, or did you mean alphabetic?
– Arnauld
Aug 14 at 1:40
 |Â
show 3 more comments
8 Answers
8
active
oldest
votes
up vote
11
down vote
Retina 0.8.2, 58 57 bytes
^((.*).)(?<=b1.*¶1)
$.2$* +
m)+`^(.*) (.*¶1[+|])
$1|$2
Try it online! Link includes one test case. Edit: Saved 1 byte thanks to @FryAmTheEggman pointing out that I overlooked a switch from b to ^ made possible by the m). Explanation:
m)
Turn on per-line ^ for the whole program.
^((.*).)(?<=^1.*¶1)
$.2$* +
For each word, try to match as much as possible from the beginning of the previous word. Change the match to spaces, except the last character, which becomes a +.
+`^(.*) (.*¶1[+|])
$1|$2
Repeatedly replace all spaces immediately above +s or |s with |s.
answered Aug 13 at 20:23
Neil
74.9k744170
@FryAmTheEggman Indeed, I added them)specifically to be able to do that, so I'm annoyed that I missed an instance.
– Neil
Aug 15 at 23:41
Ugh, why do I even bother replying to comments if people are just going to delete them...
– Neil
Aug 16 at 8:30
add a comment |Â
up vote
9
down vote
JavaScript (ES6), 128 bytes
Expects and returns a list of lists of characters.
a=>a.map((w,y)=>a[~y]=w.map(m=(c,x)=>(p=a[y-1]||0,m|=c!=p[x])?c:p[x+1]==w[x+1]?' ':(g=y=>a[y][x]<1?g(y+1,a[y][x]='|'):'+')(-y)))
Try it online!
How?
Spaces and +'s can be inserted by walking through the first to the last word in order, but |'s can only be inserted a posteriori once a + has been identified. This could be achieved by doing two distinct passes, but instead we save a pointer to each modified entry in a[~y] so that it can later be updated again within the same map() loop.
In theory, a simpler workaround would be to walk through the words in reverse order and to reverse the output as well at the end of the process. But this is a bit costly in JS and I didn't find a way to get a shorter version with this method.
a => // a = input array
a.map((w, y) => // for each word w at position y in a:
a[~y] = // save a pointer to the current entry in a[~y]
w.map(m = // initialize m to a non-numeric value
(c, x) => ( // for each character c at position x in w:
p = a[y - 1] || 0, // p = previous word or a dummy object
m |= c != p[x] // set m = 1 as soon as w differs from p at this position
) ? // if w is no longer equal to p:
c // append c
: // else:
p[x + 1] == w[x + 1] ? // if the next characters are still matching:
' ' // append a space
: ( // else:
g = y => // g() = recursive function to insert pipes
a[y][x] < 1 ? // if a[y][x] is a space:
g( // do a recursive call to g()
y + 1, // with y + 1
a[y][x] = '|' // and overwrite a[y][x] with a pipe
) // end of recursive call
: // else:
'+' // make the whole recursion chain return a '+'
// which will be appended in the current entry
)(-y) // initial call to g() with -y (this is ~y + 1)
) // end of map() over the characters
) // end of map() over the words
answered Aug 14 at 2:06
Arnauld
63.3k580267
would you look at my solution, i came up with it myself but it reminds of your solution. so if its too close you can submit it as yours (or not) and ill delete it :)
– DanielIndie
Aug 17 at 16:22
@DanielIndie No worries. It's different enough.
– Arnauld
Aug 20 at 11:00
add a comment |Â
up vote
2
down vote
JavaScript (Node.js), 125 122 118 117 bytes
a=>a.map((w,i)=>a[i]=w.map((T,k)=>+i&&l[k]==T?l[-~k]<w[-~k]?(g=_=>a[--i][k]<"+"?g(a[i][k]="|"):"+")():" ":(i=l=w,T)))
Try it online!
- thanks to @Arnauld for tio testing :)
answered Aug 17 at 16:06
DanielIndie
1,2105
add a comment |Â
up vote
1
down vote
Python, 263 260 bytes
-3 bytes thanks to Jonathan Frech
Code:
p=lambda t,f,g:"n".join([(f[:-1]+"+"if(a!=min(t))*g else"")+a+p(t[a],(f+" "if len(t[a])>1or a==max(t)else f[:-1]+"| "),1)for a in t])if t else""
def a(t,x):
if x:c=x[0];t[c]=c in t and t[c]or;a(t[c],x[1:])
def f(*s):t=;[a(t,i)for i in s];return p(t,"",0)
Try it Online!
Explanation:
This solution builds a trie out of the input words and recursively parses it into the required output. The a function takes a trie t and a string s and adds x to t. Tries are implemented as nested dictionaries. Each dictionary represents a node in the trie. For example, the dictionary representing the trie generated by the first test case looks like this:
'b': 'a': 'l': 'd': 'e': 'r': 'd': 'a': 's': 'h': , 'l': 'e': 't': , 'o': 'o': 'n': 'f': 'i': 's': 'h': , 'i': 's': 't': , 't': , 'r': 'o': 'o': 'd': 'i': 'n': 'g': , 'm':
The p function recurses through this structure and generates the string representation of the trie expected by the challenge. The f function takes a bunch of strings as arguments, adds them all to a trie with a, then returns the result of calling p on the trie.
answered Aug 14 at 2:06
Zachary Cotton
34915
1
Possible 252 bytes.
– Jonathan Frech
Aug 14 at 2:23
add a comment |Â
up vote
1
down vote
C (gcc), 165 155 bytes
Takes three arguments:
char** a: an array of null-terminated wordschar* m: an array of the length of each wordint n: the number of words in the array
f(a,m,n,i,j)char**a,*m;for(i=n;--i;)for(j=0;j<m[i]&j<m[i-1]&a[i][j]==a[i-1][j];j++)a[i][j]=a[i][j+1]^a[i-1][j+1]?43:++i<n&j<m[i]&a[i--][j]%81==43?124:32;
Try it online!
answered Aug 14 at 5:40
Curtis Bechtel
29618
1
Some quick wins
– Arnauld
Aug 14 at 10:08
@Arnauld Of course! Although isn't++i<n&j<m[i]&a[i--]undefined behavior? Can I rely on gcc evaluating it left to right?
– Curtis Bechtel
Aug 14 at 13:31
It's very likely to be undefined behavior. But we define languages by their implementation, so as long as it works consistently with this version of gcc, I think that's fine.
– Arnauld
Aug 14 at 13:34
add a comment |Â
up vote
1
down vote
Perl 6, 149 144 142 bytes
1 while s/(n.*)s(.*)$0(+o$_ Z$_).join("
")
Try it online!
I'm sure this can be golfed a more, especially as I'm not an expert on regexes. This uses much the same process as Neil's Retina answer.
answered Aug 14 at 2:40
Jo King
14.9k24084
add a comment |Â
up vote
0
down vote
Python 2, 191 bytes
def f(w,r=['']):
for b,c in zip(w[1:],w)[::-1]:
s='';d=0
for x,y,z in zip(r[0]+b,b,c+b):t=s[-1:];s=s[:-1]+[['+'*(s>'')+y,t+' |'[x in'+|']][y==z],t+y][d];d=d|(y!=z)
r=[s]+r
return[w[0]]+r
Try it online!
answered Aug 14 at 18:06
Chas Brown
4,1361319
add a comment |Â
up vote
0
down vote
Ruby, 118 bytes
->ai=1;a.maps="";a[i+=j=-1].charsc[/[
Try it online!
Accepts an array of strings, outputs by modifying the original input array in-place.
Explanation
The basic string transformation is not too complex, but in order to insert vertical pipes properly, we need to iterate in reverse order, and since reverse method is quite verbose, we'll do it in a trickier way. Here, we use map just to run the loop, leave the first word alone, and then iterate from the end using negative indices:
->a
i=1; #Initialize word indexer
a.map at word boundary with +
answered Aug 17 at 12:02
Kirill L.
2,4061116
add a comment |Â
8 Answers
8
active
oldest
votes
8 Answers
8
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
11
down vote
Retina 0.8.2, 58 57 bytes
^((.*).)(?<=b1.*¶1)
$.2$* +
m)+`^(.*) (.*¶1[+|])
$1|$2
Try it online! Link includes one test case. Edit: Saved 1 byte thanks to @FryAmTheEggman pointing out that I overlooked a switch from b to ^ made possible by the m). Explanation:
m)
Turn on per-line ^ for the whole program.
^((.*).)(?<=^1.*¶1)
$.2$* +
For each word, try to match as much as possible from the beginning of the previous word. Change the match to spaces, except the last character, which becomes a +.
+`^(.*) (.*¶1[+|])
$1|$2
Repeatedly replace all spaces immediately above +s or |s with |s.
answered Aug 13 at 20:23
Neil
74.9k744170
@FryAmTheEggman Indeed, I added them)specifically to be able to do that, so I'm annoyed that I missed an instance.
– Neil
Aug 15 at 23:41
Ugh, why do I even bother replying to comments if people are just going to delete them...
– Neil
Aug 16 at 8:30
add a comment |Â
up vote
11
down vote
Retina 0.8.2, 58 57 bytes
^((.*).)(?<=b1.*¶1)
$.2$* +
m)+`^(.*) (.*¶1[+|])
$1|$2
Try it online! Link includes one test case. Edit: Saved 1 byte thanks to @FryAmTheEggman pointing out that I overlooked a switch from b to ^ made possible by the m). Explanation:
m)
Turn on per-line ^ for the whole program.
^((.*).)(?<=^1.*¶1)
$.2$* +
For each word, try to match as much as possible from the beginning of the previous word. Change the match to spaces, except the last character, which becomes a +.
+`^(.*) (.*¶1[+|])
$1|$2
Repeatedly replace all spaces immediately above +s or |s with |s.
answered Aug 13 at 20:23
Neil
74.9k744170
@FryAmTheEggman Indeed, I added them)specifically to be able to do that, so I'm annoyed that I missed an instance.
– Neil
Aug 15 at 23:41
Ugh, why do I even bother replying to comments if people are just going to delete them...
– Neil
Aug 16 at 8:30
add a comment |Â
up vote
11
down vote
up vote
11
down vote
Retina 0.8.2, 58 57 bytes
^((.*).)(?<=b1.*¶1)
$.2$* +
m)+`^(.*) (.*¶1[+|])
$1|$2
Try it online! Link includes one test case. Edit: Saved 1 byte thanks to @FryAmTheEggman pointing out that I overlooked a switch from b to ^ made possible by the m). Explanation:
m)
Turn on per-line ^ for the whole program.
^((.*).)(?<=^1.*¶1)
$.2$* +
For each word, try to match as much as possible from the beginning of the previous word. Change the match to spaces, except the last character, which becomes a +.
+`^(.*) (.*¶1[+|])
$1|$2
Repeatedly replace all spaces immediately above +s or |s with |s.
answered Aug 13 at 20:23
Neil
74.9k744170
Retina 0.8.2, 58 57 bytes
^((.*).)(?<=b1.*¶1)
$.2$* +
m)+`^(.*) (.*¶1[+|])
$1|$2
Try it online! Link includes one test case. Edit: Saved 1 byte thanks to @FryAmTheEggman pointing out that I overlooked a switch from b to ^ made possible by the m). Explanation:
m)
Turn on per-line ^ for the whole program.
^((.*).)(?<=^1.*¶1)
$.2$* +
For each word, try to match as much as possible from the beginning of the previous word. Change the match to spaces, except the last character, which becomes a +.
+`^(.*) (.*¶1[+|])
$1|$2
Repeatedly replace all spaces immediately above +s or |s with |s.
answered Aug 13 at 20:23
Neil
74.9k744170
edited Aug 15 at 23:40
answered Aug 13 at 20:23
Neil
74.9k744170
answered Aug 13 at 20:23
Neil
74.9k744170
answered Aug 13 at 20:23
Neil
74.9k744170
74.9k744170
@FryAmTheEggman Indeed, I added them)specifically to be able to do that, so I'm annoyed that I missed an instance.
– Neil
Aug 15 at 23:41
Ugh, why do I even bother replying to comments if people are just going to delete them...
– Neil
Aug 16 at 8:30
add a comment |Â
@FryAmTheEggman Indeed, I added them)specifically to be able to do that, so I'm annoyed that I missed an instance.
– Neil
Aug 15 at 23:41
Ugh, why do I even bother replying to comments if people are just going to delete them...
– Neil
Aug 16 at 8:30
@FryAmTheEggman Indeed, I added the
m) specifically to be able to do that, so I'm annoyed that I missed an instance.– Neil
Aug 15 at 23:41
@FryAmTheEggman Indeed, I added the
m) specifically to be able to do that, so I'm annoyed that I missed an instance.– Neil
Aug 15 at 23:41
Ugh, why do I even bother replying to comments if people are just going to delete them...
– Neil
Aug 16 at 8:30
Ugh, why do I even bother replying to comments if people are just going to delete them...
– Neil
Aug 16 at 8:30
add a comment |Â
up vote
9
down vote
JavaScript (ES6), 128 bytes
Expects and returns a list of lists of characters.
a=>a.map((w,y)=>a[~y]=w.map(m=(c,x)=>(p=a[y-1]||0,m|=c!=p[x])?c:p[x+1]==w[x+1]?' ':(g=y=>a[y][x]<1?g(y+1,a[y][x]='|'):'+')(-y)))
Try it online!
How?
Spaces and +'s can be inserted by walking through the first to the last word in order, but |'s can only be inserted a posteriori once a + has been identified. This could be achieved by doing two distinct passes, but instead we save a pointer to each modified entry in a[~y] so that it can later be updated again within the same map() loop.
In theory, a simpler workaround would be to walk through the words in reverse order and to reverse the output as well at the end of the process. But this is a bit costly in JS and I didn't find a way to get a shorter version with this method.
a => // a = input array
a.map((w, y) => // for each word w at position y in a:
a[~y] = // save a pointer to the current entry in a[~y]
w.map(m = // initialize m to a non-numeric value
(c, x) => ( // for each character c at position x in w:
p = a[y - 1] || 0, // p = previous word or a dummy object
m |= c != p[x] // set m = 1 as soon as w differs from p at this position
) ? // if w is no longer equal to p:
c // append c
: // else:
p[x + 1] == w[x + 1] ? // if the next characters are still matching:
' ' // append a space
: ( // else:
g = y => // g() = recursive function to insert pipes
a[y][x] < 1 ? // if a[y][x] is a space:
g( // do a recursive call to g()
y + 1, // with y + 1
a[y][x] = '|' // and overwrite a[y][x] with a pipe
) // end of recursive call
: // else:
'+' // make the whole recursion chain return a '+'
// which will be appended in the current entry
)(-y) // initial call to g() with -y (this is ~y + 1)
) // end of map() over the characters
) // end of map() over the words
answered Aug 14 at 2:06
Arnauld
63.3k580267
would you look at my solution, i came up with it myself but it reminds of your solution. so if its too close you can submit it as yours (or not) and ill delete it :)
– DanielIndie
Aug 17 at 16:22
@DanielIndie No worries. It's different enough.
– Arnauld
Aug 20 at 11:00
add a comment |Â
up vote
9
down vote
JavaScript (ES6), 128 bytes
Expects and returns a list of lists of characters.
a=>a.map((w,y)=>a[~y]=w.map(m=(c,x)=>(p=a[y-1]||0,m|=c!=p[x])?c:p[x+1]==w[x+1]?' ':(g=y=>a[y][x]<1?g(y+1,a[y][x]='|'):'+')(-y)))
Try it online!
How?
Spaces and +'s can be inserted by walking through the first to the last word in order, but |'s can only be inserted a posteriori once a + has been identified. This could be achieved by doing two distinct passes, but instead we save a pointer to each modified entry in a[~y] so that it can later be updated again within the same map() loop.
In theory, a simpler workaround would be to walk through the words in reverse order and to reverse the output as well at the end of the process. But this is a bit costly in JS and I didn't find a way to get a shorter version with this method.
a => // a = input array
a.map((w, y) => // for each word w at position y in a:
a[~y] = // save a pointer to the current entry in a[~y]
w.map(m = // initialize m to a non-numeric value
(c, x) => ( // for each character c at position x in w:
p = a[y - 1] || 0, // p = previous word or a dummy object
m |= c != p[x] // set m = 1 as soon as w differs from p at this position
) ? // if w is no longer equal to p:
c // append c
: // else:
p[x + 1] == w[x + 1] ? // if the next characters are still matching:
' ' // append a space
: ( // else:
g = y => // g() = recursive function to insert pipes
a[y][x] < 1 ? // if a[y][x] is a space:
g( // do a recursive call to g()
y + 1, // with y + 1
a[y][x] = '|' // and overwrite a[y][x] with a pipe
) // end of recursive call
: // else:
'+' // make the whole recursion chain return a '+'
// which will be appended in the current entry
)(-y) // initial call to g() with -y (this is ~y + 1)
) // end of map() over the characters
) // end of map() over the words
answered Aug 14 at 2:06
Arnauld
63.3k580267
would you look at my solution, i came up with it myself but it reminds of your solution. so if its too close you can submit it as yours (or not) and ill delete it :)
– DanielIndie
Aug 17 at 16:22
@DanielIndie No worries. It's different enough.
– Arnauld
Aug 20 at 11:00
add a comment |Â
up vote
9
down vote
up vote
9
down vote
JavaScript (ES6), 128 bytes
Expects and returns a list of lists of characters.
a=>a.map((w,y)=>a[~y]=w.map(m=(c,x)=>(p=a[y-1]||0,m|=c!=p[x])?c:p[x+1]==w[x+1]?' ':(g=y=>a[y][x]<1?g(y+1,a[y][x]='|'):'+')(-y)))
Try it online!
How?
Spaces and +'s can be inserted by walking through the first to the last word in order, but |'s can only be inserted a posteriori once a + has been identified. This could be achieved by doing two distinct passes, but instead we save a pointer to each modified entry in a[~y] so that it can later be updated again within the same map() loop.
In theory, a simpler workaround would be to walk through the words in reverse order and to reverse the output as well at the end of the process. But this is a bit costly in JS and I didn't find a way to get a shorter version with this method.
a => // a = input array
a.map((w, y) => // for each word w at position y in a:
a[~y] = // save a pointer to the current entry in a[~y]
w.map(m = // initialize m to a non-numeric value
(c, x) => ( // for each character c at position x in w:
p = a[y - 1] || 0, // p = previous word or a dummy object
m |= c != p[x] // set m = 1 as soon as w differs from p at this position
) ? // if w is no longer equal to p:
c // append c
: // else:
p[x + 1] == w[x + 1] ? // if the next characters are still matching:
' ' // append a space
: ( // else:
g = y => // g() = recursive function to insert pipes
a[y][x] < 1 ? // if a[y][x] is a space:
g( // do a recursive call to g()
y + 1, // with y + 1
a[y][x] = '|' // and overwrite a[y][x] with a pipe
) // end of recursive call
: // else:
'+' // make the whole recursion chain return a '+'
// which will be appended in the current entry
)(-y) // initial call to g() with -y (this is ~y + 1)
) // end of map() over the characters
) // end of map() over the words
answered Aug 14 at 2:06
Arnauld
63.3k580267
JavaScript (ES6), 128 bytes
Expects and returns a list of lists of characters.
a=>a.map((w,y)=>a[~y]=w.map(m=(c,x)=>(p=a[y-1]||0,m|=c!=p[x])?c:p[x+1]==w[x+1]?' ':(g=y=>a[y][x]<1?g(y+1,a[y][x]='|'):'+')(-y)))
Try it online!
How?
Spaces and +'s can be inserted by walking through the first to the last word in order, but |'s can only be inserted a posteriori once a + has been identified. This could be achieved by doing two distinct passes, but instead we save a pointer to each modified entry in a[~y] so that it can later be updated again within the same map() loop.
In theory, a simpler workaround would be to walk through the words in reverse order and to reverse the output as well at the end of the process. But this is a bit costly in JS and I didn't find a way to get a shorter version with this method.
a => // a = input array
a.map((w, y) => // for each word w at position y in a:
a[~y] = // save a pointer to the current entry in a[~y]
w.map(m = // initialize m to a non-numeric value
(c, x) => ( // for each character c at position x in w:
p = a[y - 1] || 0, // p = previous word or a dummy object
m |= c != p[x] // set m = 1 as soon as w differs from p at this position
) ? // if w is no longer equal to p:
c // append c
: // else:
p[x + 1] == w[x + 1] ? // if the next characters are still matching:
' ' // append a space
: ( // else:
g = y => // g() = recursive function to insert pipes
a[y][x] < 1 ? // if a[y][x] is a space:
g( // do a recursive call to g()
y + 1, // with y + 1
a[y][x] = '|' // and overwrite a[y][x] with a pipe
) // end of recursive call
: // else:
'+' // make the whole recursion chain return a '+'
// which will be appended in the current entry
)(-y) // initial call to g() with -y (this is ~y + 1)
) // end of map() over the characters
) // end of map() over the words
answered Aug 14 at 2:06
Arnauld
63.3k580267
edited Aug 14 at 13:50
answered Aug 14 at 2:06
Arnauld
63.3k580267
answered Aug 14 at 2:06
Arnauld
63.3k580267
answered Aug 14 at 2:06
Arnauld
63.3k580267
63.3k580267
would you look at my solution, i came up with it myself but it reminds of your solution. so if its too close you can submit it as yours (or not) and ill delete it :)
– DanielIndie
Aug 17 at 16:22
@DanielIndie No worries. It's different enough.
– Arnauld
Aug 20 at 11:00
add a comment |Â
would you look at my solution, i came up with it myself but it reminds of your solution. so if its too close you can submit it as yours (or not) and ill delete it :)
– DanielIndie
Aug 17 at 16:22
@DanielIndie No worries. It's different enough.
– Arnauld
Aug 20 at 11:00
would you look at my solution, i came up with it myself but it reminds of your solution. so if its too close you can submit it as yours (or not) and ill delete it :)
– DanielIndie
Aug 17 at 16:22
would you look at my solution, i came up with it myself but it reminds of your solution. so if its too close you can submit it as yours (or not) and ill delete it :)
– DanielIndie
Aug 17 at 16:22
@DanielIndie No worries. It's different enough.
– Arnauld
Aug 20 at 11:00
@DanielIndie No worries. It's different enough.
– Arnauld
Aug 20 at 11:00
add a comment |Â
up vote
2
down vote
JavaScript (Node.js), 125 122 118 117 bytes
a=>a.map((w,i)=>a[i]=w.map((T,k)=>+i&&l[k]==T?l[-~k]<w[-~k]?(g=_=>a[--i][k]<"+"?g(a[i][k]="|"):"+")():" ":(i=l=w,T)))
Try it online!
- thanks to @Arnauld for tio testing :)
answered Aug 17 at 16:06
DanielIndie
1,2105
add a comment |Â
up vote
2
down vote
JavaScript (Node.js), 125 122 118 117 bytes
a=>a.map((w,i)=>a[i]=w.map((T,k)=>+i&&l[k]==T?l[-~k]<w[-~k]?(g=_=>a[--i][k]<"+"?g(a[i][k]="|"):"+")():" ":(i=l=w,T)))
Try it online!
- thanks to @Arnauld for tio testing :)
answered Aug 17 at 16:06
DanielIndie
1,2105
add a comment |Â
up vote
2
down vote
up vote
2
down vote
JavaScript (Node.js), 125 122 118 117 bytes
a=>a.map((w,i)=>a[i]=w.map((T,k)=>+i&&l[k]==T?l[-~k]<w[-~k]?(g=_=>a[--i][k]<"+"?g(a[i][k]="|"):"+")():" ":(i=l=w,T)))
Try it online!
- thanks to @Arnauld for tio testing :)
answered Aug 17 at 16:06
DanielIndie
1,2105
JavaScript (Node.js), 125 122 118 117 bytes
a=>a.map((w,i)=>a[i]=w.map((T,k)=>+i&&l[k]==T?l[-~k]<w[-~k]?(g=_=>a[--i][k]<"+"?g(a[i][k]="|"):"+")():" ":(i=l=w,T)))
Try it online!
- thanks to @Arnauld for tio testing :)
answered Aug 17 at 16:06
DanielIndie
1,2105
edited Aug 17 at 19:23
answered Aug 17 at 16:06
DanielIndie
1,2105
answered Aug 17 at 16:06
DanielIndie
1,2105
answered Aug 17 at 16:06
DanielIndie
1,2105
1,2105
add a comment |Â
add a comment |Â
up vote
1
down vote
Python, 263 260 bytes
-3 bytes thanks to Jonathan Frech
Code:
p=lambda t,f,g:"n".join([(f[:-1]+"+"if(a!=min(t))*g else"")+a+p(t[a],(f+" "if len(t[a])>1or a==max(t)else f[:-1]+"| "),1)for a in t])if t else""
def a(t,x):
if x:c=x[0];t[c]=c in t and t[c]or;a(t[c],x[1:])
def f(*s):t=;[a(t,i)for i in s];return p(t,"",0)
Try it Online!
Explanation:
This solution builds a trie out of the input words and recursively parses it into the required output. The a function takes a trie t and a string s and adds x to t. Tries are implemented as nested dictionaries. Each dictionary represents a node in the trie. For example, the dictionary representing the trie generated by the first test case looks like this:
'b': 'a': 'l': 'd': 'e': 'r': 'd': 'a': 's': 'h': , 'l': 'e': 't': , 'o': 'o': 'n': 'f': 'i': 's': 'h': , 'i': 's': 't': , 't': , 'r': 'o': 'o': 'd': 'i': 'n': 'g': , 'm':
The p function recurses through this structure and generates the string representation of the trie expected by the challenge. The f function takes a bunch of strings as arguments, adds them all to a trie with a, then returns the result of calling p on the trie.
answered Aug 14 at 2:06
Zachary Cotton
34915
1
Possible 252 bytes.
– Jonathan Frech
Aug 14 at 2:23
add a comment |Â
up vote
1
down vote
Python, 263 260 bytes
-3 bytes thanks to Jonathan Frech
Code:
p=lambda t,f,g:"n".join([(f[:-1]+"+"if(a!=min(t))*g else"")+a+p(t[a],(f+" "if len(t[a])>1or a==max(t)else f[:-1]+"| "),1)for a in t])if t else""
def a(t,x):
if x:c=x[0];t[c]=c in t and t[c]or;a(t[c],x[1:])
def f(*s):t=;[a(t,i)for i in s];return p(t,"",0)
Try it Online!
Explanation:
This solution builds a trie out of the input words and recursively parses it into the required output. The a function takes a trie t and a string s and adds x to t. Tries are implemented as nested dictionaries. Each dictionary represents a node in the trie. For example, the dictionary representing the trie generated by the first test case looks like this:
'b': 'a': 'l': 'd': 'e': 'r': 'd': 'a': 's': 'h': , 'l': 'e': 't': , 'o': 'o': 'n': 'f': 'i': 's': 'h': , 'i': 's': 't': , 't': , 'r': 'o': 'o': 'd': 'i': 'n': 'g': , 'm':
The p function recurses through this structure and generates the string representation of the trie expected by the challenge. The f function takes a bunch of strings as arguments, adds them all to a trie with a, then returns the result of calling p on the trie.
answered Aug 14 at 2:06
Zachary Cotton
34915
1
Possible 252 bytes.
– Jonathan Frech
Aug 14 at 2:23
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Python, 263 260 bytes
-3 bytes thanks to Jonathan Frech
Code:
p=lambda t,f,g:"n".join([(f[:-1]+"+"if(a!=min(t))*g else"")+a+p(t[a],(f+" "if len(t[a])>1or a==max(t)else f[:-1]+"| "),1)for a in t])if t else""
def a(t,x):
if x:c=x[0];t[c]=c in t and t[c]or;a(t[c],x[1:])
def f(*s):t=;[a(t,i)for i in s];return p(t,"",0)
Try it Online!
Explanation:
This solution builds a trie out of the input words and recursively parses it into the required output. The a function takes a trie t and a string s and adds x to t. Tries are implemented as nested dictionaries. Each dictionary represents a node in the trie. For example, the dictionary representing the trie generated by the first test case looks like this:
'b': 'a': 'l': 'd': 'e': 'r': 'd': 'a': 's': 'h': , 'l': 'e': 't': , 'o': 'o': 'n': 'f': 'i': 's': 'h': , 'i': 's': 't': , 't': , 'r': 'o': 'o': 'd': 'i': 'n': 'g': , 'm':
The p function recurses through this structure and generates the string representation of the trie expected by the challenge. The f function takes a bunch of strings as arguments, adds them all to a trie with a, then returns the result of calling p on the trie.
answered Aug 14 at 2:06
Zachary Cotton
34915
Python, 263 260 bytes
-3 bytes thanks to Jonathan Frech
Code:
p=lambda t,f,g:"n".join([(f[:-1]+"+"if(a!=min(t))*g else"")+a+p(t[a],(f+" "if len(t[a])>1or a==max(t)else f[:-1]+"| "),1)for a in t])if t else""
def a(t,x):
if x:c=x[0];t[c]=c in t and t[c]or;a(t[c],x[1:])
def f(*s):t=;[a(t,i)for i in s];return p(t,"",0)
Try it Online!
Explanation:
This solution builds a trie out of the input words and recursively parses it into the required output. The a function takes a trie t and a string s and adds x to t. Tries are implemented as nested dictionaries. Each dictionary represents a node in the trie. For example, the dictionary representing the trie generated by the first test case looks like this:
'b': 'a': 'l': 'd': 'e': 'r': 'd': 'a': 's': 'h': , 'l': 'e': 't': , 'o': 'o': 'n': 'f': 'i': 's': 'h': , 'i': 's': 't': , 't': , 'r': 'o': 'o': 'd': 'i': 'n': 'g': , 'm':
The p function recurses through this structure and generates the string representation of the trie expected by the challenge. The f function takes a bunch of strings as arguments, adds them all to a trie with a, then returns the result of calling p on the trie.
answered Aug 14 at 2:06
Zachary Cotton
34915
edited Aug 14 at 2:27
answered Aug 14 at 2:06
Zachary Cotton
34915
answered Aug 14 at 2:06
Zachary Cotton
34915
answered Aug 14 at 2:06
Zachary Cotton
34915
34915
1
Possible 252 bytes.
– Jonathan Frech
Aug 14 at 2:23
add a comment |Â
1
Possible 252 bytes.
– Jonathan Frech
Aug 14 at 2:23
1
1
Possible 252 bytes.
– Jonathan Frech
Aug 14 at 2:23
Possible 252 bytes.
– Jonathan Frech
Aug 14 at 2:23
add a comment |Â
up vote
1
down vote
C (gcc), 165 155 bytes
Takes three arguments:
char** a: an array of null-terminated wordschar* m: an array of the length of each wordint n: the number of words in the array
f(a,m,n,i,j)char**a,*m;for(i=n;--i;)for(j=0;j<m[i]&j<m[i-1]&a[i][j]==a[i-1][j];j++)a[i][j]=a[i][j+1]^a[i-1][j+1]?43:++i<n&j<m[i]&a[i--][j]%81==43?124:32;
Try it online!
answered Aug 14 at 5:40
Curtis Bechtel
29618
1
Some quick wins
– Arnauld
Aug 14 at 10:08
@Arnauld Of course! Although isn't++i<n&j<m[i]&a[i--]undefined behavior? Can I rely on gcc evaluating it left to right?
– Curtis Bechtel
Aug 14 at 13:31
It's very likely to be undefined behavior. But we define languages by their implementation, so as long as it works consistently with this version of gcc, I think that's fine.
– Arnauld
Aug 14 at 13:34
add a comment |Â
up vote
1
down vote
C (gcc), 165 155 bytes
Takes three arguments:
char** a: an array of null-terminated wordschar* m: an array of the length of each wordint n: the number of words in the array
f(a,m,n,i,j)char**a,*m;for(i=n;--i;)for(j=0;j<m[i]&j<m[i-1]&a[i][j]==a[i-1][j];j++)a[i][j]=a[i][j+1]^a[i-1][j+1]?43:++i<n&j<m[i]&a[i--][j]%81==43?124:32;
Try it online!
answered Aug 14 at 5:40
Curtis Bechtel
29618
1
Some quick wins
– Arnauld
Aug 14 at 10:08
@Arnauld Of course! Although isn't++i<n&j<m[i]&a[i--]undefined behavior? Can I rely on gcc evaluating it left to right?
– Curtis Bechtel
Aug 14 at 13:31
It's very likely to be undefined behavior. But we define languages by their implementation, so as long as it works consistently with this version of gcc, I think that's fine.
– Arnauld
Aug 14 at 13:34
add a comment |Â
up vote
1
down vote
up vote
1
down vote
C (gcc), 165 155 bytes
Takes three arguments:
char** a: an array of null-terminated wordschar* m: an array of the length of each wordint n: the number of words in the array
f(a,m,n,i,j)char**a,*m;for(i=n;--i;)for(j=0;j<m[i]&j<m[i-1]&a[i][j]==a[i-1][j];j++)a[i][j]=a[i][j+1]^a[i-1][j+1]?43:++i<n&j<m[i]&a[i--][j]%81==43?124:32;
Try it online!
answered Aug 14 at 5:40
Curtis Bechtel
29618
C (gcc), 165 155 bytes
Takes three arguments:
char** a: an array of null-terminated wordschar* m: an array of the length of each wordint n: the number of words in the array
f(a,m,n,i,j)char**a,*m;for(i=n;--i;)for(j=0;j<m[i]&j<m[i-1]&a[i][j]==a[i-1][j];j++)a[i][j]=a[i][j+1]^a[i-1][j+1]?43:++i<n&j<m[i]&a[i--][j]%81==43?124:32;
Try it online!
answered Aug 14 at 5:40
Curtis Bechtel
29618
edited Aug 14 at 13:38
answered Aug 14 at 5:40
Curtis Bechtel
29618
answered Aug 14 at 5:40
Curtis Bechtel
29618
answered Aug 14 at 5:40
Curtis Bechtel
29618
29618
1
Some quick wins
– Arnauld
Aug 14 at 10:08
@Arnauld Of course! Although isn't++i<n&j<m[i]&a[i--]undefined behavior? Can I rely on gcc evaluating it left to right?
– Curtis Bechtel
Aug 14 at 13:31
It's very likely to be undefined behavior. But we define languages by their implementation, so as long as it works consistently with this version of gcc, I think that's fine.
– Arnauld
Aug 14 at 13:34
add a comment |Â
1
Some quick wins
– Arnauld
Aug 14 at 10:08
@Arnauld Of course! Although isn't++i<n&j<m[i]&a[i--]undefined behavior? Can I rely on gcc evaluating it left to right?
– Curtis Bechtel
Aug 14 at 13:31
It's very likely to be undefined behavior. But we define languages by their implementation, so as long as it works consistently with this version of gcc, I think that's fine.
– Arnauld
Aug 14 at 13:34
1
1
Some quick wins
– Arnauld
Aug 14 at 10:08
Some quick wins
– Arnauld
Aug 14 at 10:08
@Arnauld Of course! Although isn't
++i<n&j<m[i]&a[i--] undefined behavior? Can I rely on gcc evaluating it left to right?– Curtis Bechtel
Aug 14 at 13:31
@Arnauld Of course! Although isn't
++i<n&j<m[i]&a[i--] undefined behavior? Can I rely on gcc evaluating it left to right?– Curtis Bechtel
Aug 14 at 13:31
It's very likely to be undefined behavior. But we define languages by their implementation, so as long as it works consistently with this version of gcc, I think that's fine.
– Arnauld
Aug 14 at 13:34
It's very likely to be undefined behavior. But we define languages by their implementation, so as long as it works consistently with this version of gcc, I think that's fine.
– Arnauld
Aug 14 at 13:34
add a comment |Â
up vote
1
down vote
Perl 6, 149 144 142 bytes
1 while s/(n.*)s(.*)$0(+o$_ Z$_).join("
")
Try it online!
I'm sure this can be golfed a more, especially as I'm not an expert on regexes. This uses much the same process as Neil's Retina answer.
answered Aug 14 at 2:40
Jo King
14.9k24084
add a comment |Â
up vote
1
down vote
Perl 6, 149 144 142 bytes
1 while s/(n.*)s(.*)$0(+o$_ Z$_).join("
")
Try it online!
I'm sure this can be golfed a more, especially as I'm not an expert on regexes. This uses much the same process as Neil's Retina answer.
answered Aug 14 at 2:40
Jo King
14.9k24084
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Perl 6, 149 144 142 bytes
1 while s/(n.*)s(.*)$0(+o$_ Z$_).join("
")
Try it online!
I'm sure this can be golfed a more, especially as I'm not an expert on regexes. This uses much the same process as Neil's Retina answer.
answered Aug 14 at 2:40
Jo King
14.9k24084
Perl 6, 149 144 142 bytes
1 while s/(n.*)s(.*)$0(+o$_ Z$_).join("
")
Try it online!
I'm sure this can be golfed a more, especially as I'm not an expert on regexes. This uses much the same process as Neil's Retina answer.
answered Aug 14 at 2:40
Jo King
14.9k24084
edited Aug 17 at 12:30
answered Aug 14 at 2:40
Jo King
14.9k24084
answered Aug 14 at 2:40
Jo King
14.9k24084
answered Aug 14 at 2:40
Jo King
14.9k24084
14.9k24084
add a comment |Â
add a comment |Â
up vote
0
down vote
Python 2, 191 bytes
def f(w,r=['']):
for b,c in zip(w[1:],w)[::-1]:
s='';d=0
for x,y,z in zip(r[0]+b,b,c+b):t=s[-1:];s=s[:-1]+[['+'*(s>'')+y,t+' |'[x in'+|']][y==z],t+y][d];d=d|(y!=z)
r=[s]+r
return[w[0]]+r
Try it online!
answered Aug 14 at 18:06
Chas Brown
4,1361319
add a comment |Â
up vote
0
down vote
Python 2, 191 bytes
def f(w,r=['']):
for b,c in zip(w[1:],w)[::-1]:
s='';d=0
for x,y,z in zip(r[0]+b,b,c+b):t=s[-1:];s=s[:-1]+[['+'*(s>'')+y,t+' |'[x in'+|']][y==z],t+y][d];d=d|(y!=z)
r=[s]+r
return[w[0]]+r
Try it online!
answered Aug 14 at 18:06
Chas Brown
4,1361319
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Python 2, 191 bytes
def f(w,r=['']):
for b,c in zip(w[1:],w)[::-1]:
s='';d=0
for x,y,z in zip(r[0]+b,b,c+b):t=s[-1:];s=s[:-1]+[['+'*(s>'')+y,t+' |'[x in'+|']][y==z],t+y][d];d=d|(y!=z)
r=[s]+r
return[w[0]]+r
Try it online!
answered Aug 14 at 18:06
Chas Brown
4,1361319
Python 2, 191 bytes
def f(w,r=['']):
for b,c in zip(w[1:],w)[::-1]:
s='';d=0
for x,y,z in zip(r[0]+b,b,c+b):t=s[-1:];s=s[:-1]+[['+'*(s>'')+y,t+' |'[x in'+|']][y==z],t+y][d];d=d|(y!=z)
r=[s]+r
return[w[0]]+r
Try it online!
answered Aug 14 at 18:06
Chas Brown
4,1361319
answered Aug 14 at 18:06
Chas Brown
4,1361319
answered Aug 14 at 18:06
Chas Brown
4,1361319
answered Aug 14 at 18:06
Chas Brown
4,1361319
4,1361319
add a comment |Â
add a comment |Â
up vote
0
down vote
Ruby, 118 bytes
->ai=1;a.maps="";a[i+=j=-1].charsc[/[
Try it online!
Accepts an array of strings, outputs by modifying the original input array in-place.
Explanation
The basic string transformation is not too complex, but in order to insert vertical pipes properly, we need to iterate in reverse order, and since reverse method is quite verbose, we'll do it in a trickier way. Here, we use map just to run the loop, leave the first word alone, and then iterate from the end using negative indices:
->a
i=1; #Initialize word indexer
a.map at word boundary with +
answered Aug 17 at 12:02
Kirill L.
2,4061116
add a comment |Â
up vote
0
down vote
Ruby, 118 bytes
->ai=1;a.maps="";a[i+=j=-1].charsc[/[
Try it online!
Accepts an array of strings, outputs by modifying the original input array in-place.
Explanation
The basic string transformation is not too complex, but in order to insert vertical pipes properly, we need to iterate in reverse order, and since reverse method is quite verbose, we'll do it in a trickier way. Here, we use map just to run the loop, leave the first word alone, and then iterate from the end using negative indices:
->a
i=1; #Initialize word indexer
a.map at word boundary with +
answered Aug 17 at 12:02
Kirill L.
2,4061116
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Ruby, 118 bytes
->ai=1;a.maps="";a[i+=j=-1].charsc[/[
Try it online!
Accepts an array of strings, outputs by modifying the original input array in-place.
Explanation
The basic string transformation is not too complex, but in order to insert vertical pipes properly, we need to iterate in reverse order, and since reverse method is quite verbose, we'll do it in a trickier way. Here, we use map just to run the loop, leave the first word alone, and then iterate from the end using negative indices:
->a
i=1; #Initialize word indexer
a.map at word boundary with +
answered Aug 17 at 12:02
Kirill L.
2,4061116
Ruby, 118 bytes
->ai=1;a.maps="";a[i+=j=-1].charsc[/[
Try it online!
Accepts an array of strings, outputs by modifying the original input array in-place.
Explanation
The basic string transformation is not too complex, but in order to insert vertical pipes properly, we need to iterate in reverse order, and since reverse method is quite verbose, we'll do it in a trickier way. Here, we use map just to run the loop, leave the first word alone, and then iterate from the end using negative indices:
->a
i=1; #Initialize word indexer
a.map at word boundary with +
answered Aug 17 at 12:02
Kirill L.
2,4061116
edited Aug 17 at 13:56
answered Aug 17 at 12:02
Kirill L.
2,4061116
answered Aug 17 at 12:02
Kirill L.
2,4061116
answered Aug 17 at 12:02
Kirill L.
2,4061116
2,4061116
add a comment |Â
add a comment |Â
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What about the case where I have the word
ballafterballoon. What output should we expect?– Rushabh Mehta
Aug 13 at 20:05
@RushabhMehta I'm guessing you would just have a
+under the firsto, but I didn't write the challenge so I'm not certain.– Theo
Aug 13 at 20:12
5
@RushabhMehta The words are alphabetically sorted, so this won't happen.
– Neil
Aug 13 at 20:14
@Neil Oh good point
– Rushabh Mehta
Aug 13 at 20:18
2
The words in the input will be made up of only alphanumerics: does that really include digits, or did you mean alphabetic?
– Arnauld
Aug 14 at 1:40