Same number letters

Clash Royale CLAN TAG#URR8PPP

up vote
2
down vote

favorite

Letters of the words want fairness

they decided to appear same number of times in a sentence equally

Example

Priorities

will become

Ppprrioooritttieeesss

each letter show 3 times as letter mostly used was i and it showed 3 times

it does not matter where you put the repeated letters as log as they next to a similar letter

means

Pppriooorritttieeesss is ok (the r letter)

Ppprioororitttieeesss is no ok (the r letter)

another example

invoice

will become

innvvooiccee

another example

Remittance Advice

will become

Rrremmmiitttaannncce Adddvvvice

Space, comma, question mark, quotation, etc. are not letters letters are from (a-z both cap and small) , just once space is enough, and the order of the letters should remain the same

the form of the letter does not matter, either cap or small will count as the same letter, Pip has 2 Ps and 1 I so it will become Piip

It is case insensitive letters can be in any form, Piip=piip=piiP=PiiP

This is code-golf

code-golf word-puzzle

share|improve this question

edited 2 mins ago

asked 5 hours ago

asmgx

1156

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  How do we deal with capital letters?
  – xnor
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Caps and small no difference, you can use either
  – asmgx
  4 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Must the original (relative) order of the letters remain? If not an algorithm like the Perl answer would dominate.
  – Shieru Asakoto
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @JonathanAllan will fix this now
  – asmgx
  11 mins ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  can we change the case of capitalised letters? e.g. Can Pip result in piip?
  – Jo King
  7 mins ago
  

  
  
  
  

 | 
show 6 more comments

up vote
2
down vote

favorite

Letters of the words want fairness

they decided to appear same number of times in a sentence equally

Example

Priorities

will become

Ppprrioooritttieeesss

each letter show 3 times as letter mostly used was i and it showed 3 times

it does not matter where you put the repeated letters as log as they next to a similar letter

means

Pppriooorritttieeesss is ok (the r letter)

Ppprioororitttieeesss is no ok (the r letter)

another example

invoice

will become

innvvooiccee

another example

Remittance Advice

will become

Rrremmmiitttaannncce Adddvvvice

Space, comma, question mark, quotation, etc. are not letters letters are from (a-z both cap and small) , just once space is enough, and the order of the letters should remain the same

the form of the letter does not matter, either cap or small will count as the same letter, Pip has 2 Ps and 1 I so it will become Piip

It is case insensitive letters can be in any form, Piip=piip=piiP=PiiP

This is code-golf

code-golf word-puzzle

share|improve this question

edited 2 mins ago

asked 5 hours ago

asmgx

1156

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  How do we deal with capital letters?
  – xnor
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Caps and small no difference, you can use either
  – asmgx
  4 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Must the original (relative) order of the letters remain? If not an algorithm like the Perl answer would dominate.
  – Shieru Asakoto
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @JonathanAllan will fix this now
  – asmgx
  11 mins ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  can we change the case of capitalised letters? e.g. Can Pip result in piip?
  – Jo King
  7 mins ago
  

  
  
  
  

 | 
show 6 more comments

up vote
2
down vote

favorite

up vote
2
down vote

favorite

Letters of the words want fairness

they decided to appear same number of times in a sentence equally

Example

Priorities

will become

Ppprrioooritttieeesss

each letter show 3 times as letter mostly used was i and it showed 3 times

it does not matter where you put the repeated letters as log as they next to a similar letter

means

Pppriooorritttieeesss is ok (the r letter)

Ppprioororitttieeesss is no ok (the r letter)

another example

invoice

will become

innvvooiccee

another example

Remittance Advice

will become

Rrremmmiitttaannncce Adddvvvice

Space, comma, question mark, quotation, etc. are not letters letters are from (a-z both cap and small) , just once space is enough, and the order of the letters should remain the same

the form of the letter does not matter, either cap or small will count as the same letter, Pip has 2 Ps and 1 I so it will become Piip

It is case insensitive letters can be in any form, Piip=piip=piiP=PiiP

This is code-golf

code-golf word-puzzle

share|improve this question

edited 2 mins ago

asked 5 hours ago

asmgx

1156

Letters of the words want fairness

they decided to appear same number of times in a sentence equally

Example

Priorities

will become

Ppprrioooritttieeesss

each letter show 3 times as letter mostly used was i and it showed 3 times

it does not matter where you put the repeated letters as log as they next to a similar letter

means

Pppriooorritttieeesss is ok (the r letter)

Ppprioororitttieeesss is no ok (the r letter)

another example

invoice

will become

innvvooiccee

another example

Remittance Advice

will become

Rrremmmiitttaannncce Adddvvvice

Space, comma, question mark, quotation, etc. are not letters letters are from (a-z both cap and small) , just once space is enough, and the order of the letters should remain the same

the form of the letter does not matter, either cap or small will count as the same letter, Pip has 2 Ps and 1 I so it will become Piip

It is case insensitive letters can be in any form, Piip=piip=piiP=PiiP

This is code-golf

code-golf word-puzzle

code-golf word-puzzle

share|improve this question

edited 2 mins ago

asked 5 hours ago

asmgx

1156

share|improve this question

edited 2 mins ago

asked 5 hours ago

asmgx

1156

share|improve this question

share|improve this question

edited 2 mins ago

edited 2 mins ago

edited 2 mins ago

asked 5 hours ago

asmgx

1156

asked 5 hours ago

asmgx

1156

asked 5 hours ago

asmgx

1156

1156

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  How do we deal with capital letters?
  – xnor
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Caps and small no difference, you can use either
  – asmgx
  4 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Must the original (relative) order of the letters remain? If not an algorithm like the Perl answer would dominate.
  – Shieru Asakoto
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @JonathanAllan will fix this now
  – asmgx
  11 mins ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  can we change the case of capitalised letters? e.g. Can Pip result in piip?
  – Jo King
  7 mins ago
  

  
  
  
  

 | 
show 6 more comments

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  How do we deal with capital letters?
  – xnor
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Caps and small no difference, you can use either
  – asmgx
  4 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Must the original (relative) order of the letters remain? If not an algorithm like the Perl answer would dominate.
  – Shieru Asakoto
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @JonathanAllan will fix this now
  – asmgx
  11 mins ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  can we change the case of capitalised letters? e.g. Can Pip result in piip?
  – Jo King
  7 mins ago
  

  
  
  
  

1

1

How do we deal with capital letters?
– xnor
4 hours ago

How do we deal with capital letters?
– xnor
4 hours ago

Caps and small no difference, you can use either
– asmgx
4 hours ago

Caps and small no difference, you can use either
– asmgx
4 hours ago

1

1

Must the original (relative) order of the letters remain? If not an algorithm like the Perl answer would dominate.
– Shieru Asakoto
4 hours ago

Must the original (relative) order of the letters remain? If not an algorithm like the Perl answer would dominate.
– Shieru Asakoto
4 hours ago

1

1

@JonathanAllan will fix this now
– asmgx
11 mins ago

@JonathanAllan will fix this now
– asmgx
11 mins ago

2

2

can we change the case of capitalised letters? e.g. Can Pip result in piip?
– Jo King
7 mins ago

can we change the case of capitalised letters? e.g. Can Pip result in piip?
– Jo King
7 mins ago

 | 
show 6 more comments

4 Answers
4

active

oldest

votes

up vote
3
down vote

05AB1E, 18 13 bytes

Case sensitive.

áÙSйs¢Zα>×.;

Try it online!

Explanation

á # keep only letters of input
 Ù # remove duplicates
 S # split to list of chars
 Ð # triplicate
 ¹s¢ # count the occurrence of each unique letter in input
 Zα # take abs with the max occurrence
 > # increment
 × # repeat each unique letter that many times
 .; # replace the first occurrence of the letter in the input with this

share|improve this answer

edited 17 mins ago

answered 29 mins ago

Emigna

43.5k431131

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The input is meant to be able to contain mixed case, then be treated case insensitively e.g. Pip -> Ppii; so this does not currently fit spec. However as Jo king just asked we do not know if ppii or PPII would be acceptable...
  – Jonathan Allan
  5 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @JonathanAllan: Yeah, I'm following the comments to see how the specs are finalized.
  – Emigna
  1 min ago
  

  
  
  
  

add a comment | 

up vote
2
down vote

JavaScript (Node.js), 126 bytes

f=x=>[...x].map(c=>(f[c]=-~f[c],f[c]>w?w=f[c]:w,c),w=0).map(c=>x=x.replace(c,c.repeat(parseInt(c,36)>9?w-f[c]+1:1),f[c]=w))&&x

Try it online!

all lower or ALL UPPER only.

Explanation

f = x => // The function. `f=` is needed because we will
 // use the function object as temporary storage
 [...x].map(c => ( // - Iterate among each character...
 f[c] = -~f[c], // - Add one to the character counter
 f[c] > w ? w = f[c] : w, // - Update the maximum count if necessary
 c // - Return back the character for the use in
 // the next map function
 ), w = 0) // - The max counter
 .map(c => // - Iterate again...
 x = x.replace( // - Repeat the first appearance of
 c, // - Each character
 c.repeat( // - Needed number times
 parseInt(c, 36) > 9 // - Additional constraint
 ? w - f[c] + 1 // - If this is letter, repeat
 : 1 // - If not, stay as is
 ), // - That should've been clearly stated
 f[c] = w // - And set the counter so that no further 
 // replacements are done on this character 
 ) // - (w - f[c] + 1 = 1 in further iterations)
 ) && x // - Return the result

share|improve this answer

edited 45 mins ago

answered 4 hours ago

Shieru Asakoto

1,690311

add a comment | 

up vote
2
down vote

J, 44 bytes

#~1+(</"1@=*a)+/@(*>./-])#/.~*a=.1='`z'I.~.

Try it online!

share|improve this answer

edited 15 mins ago

answered 3 hours ago

FrownyFrog

2,1671518

add a comment | 

up vote
1
down vote

Perl 6, 79 bytes

->aa.=subst($_,$_ x a.comb(/<:Ll>/).Bag.values.max+1-a.comb($_))for 'a'..'z'

Try it online!

Takes a mutable string and edits it in place. This makes no distinction between case, so input should be either all upper or all lower case. Oof, this feels like it could be shorter.

Explanation:

->a # Anonymous code block that takes a mutable string 
 for 'a'..'z' # For each letter
 a.=subst( ) # Substitute
 $_, #The first occurrence of the letter with
 $_ x #The letter repeated
 a.comb(/<:Ll>/) The lowercase letters of the string
 .Bag.values.max # The count of the most common letter
 +1 # Plus 1
 -a.comb($_) # Minus the count of that letter already in the string

share|improve this answer

edited 19 mins ago

answered 4 hours ago

Jo King

16.8k24193

add a comment | 

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4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote

05AB1E, 18 13 bytes

Case sensitive.

áÙSйs¢Zα>×.;

Try it online!

Explanation

á # keep only letters of input
 Ù # remove duplicates
 S # split to list of chars
 Ð # triplicate
 ¹s¢ # count the occurrence of each unique letter in input
 Zα # take abs with the max occurrence
 > # increment
 × # repeat each unique letter that many times
 .; # replace the first occurrence of the letter in the input with this

share|improve this answer

edited 17 mins ago

answered 29 mins ago

Emigna

43.5k431131

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The input is meant to be able to contain mixed case, then be treated case insensitively e.g. Pip -> Ppii; so this does not currently fit spec. However as Jo king just asked we do not know if ppii or PPII would be acceptable...
  – Jonathan Allan
  5 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @JonathanAllan: Yeah, I'm following the comments to see how the specs are finalized.
  – Emigna
  1 min ago
  

  
  
  
  

add a comment | 

up vote
3
down vote

05AB1E, 18 13 bytes

Case sensitive.

áÙSйs¢Zα>×.;

Try it online!

Explanation

á # keep only letters of input
 Ù # remove duplicates
 S # split to list of chars
 Ð # triplicate
 ¹s¢ # count the occurrence of each unique letter in input
 Zα # take abs with the max occurrence
 > # increment
 × # repeat each unique letter that many times
 .; # replace the first occurrence of the letter in the input with this

share|improve this answer

edited 17 mins ago

answered 29 mins ago

Emigna

43.5k431131

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The input is meant to be able to contain mixed case, then be treated case insensitively e.g. Pip -> Ppii; so this does not currently fit spec. However as Jo king just asked we do not know if ppii or PPII would be acceptable...
  – Jonathan Allan
  5 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @JonathanAllan: Yeah, I'm following the comments to see how the specs are finalized.
  – Emigna
  1 min ago
  

  
  
  
  

add a comment | 

up vote
3
down vote

up vote
3
down vote

05AB1E, 18 13 bytes

Case sensitive.

áÙSйs¢Zα>×.;

Try it online!

Explanation

á # keep only letters of input
 Ù # remove duplicates
 S # split to list of chars
 Ð # triplicate
 ¹s¢ # count the occurrence of each unique letter in input
 Zα # take abs with the max occurrence
 > # increment
 × # repeat each unique letter that many times
 .; # replace the first occurrence of the letter in the input with this

share|improve this answer

edited 17 mins ago

answered 29 mins ago

Emigna

43.5k431131

05AB1E, 18 13 bytes

Case sensitive.

áÙSйs¢Zα>×.;

Try it online!

Explanation

á # keep only letters of input
 Ù # remove duplicates
 S # split to list of chars
 Ð # triplicate
 ¹s¢ # count the occurrence of each unique letter in input
 Zα # take abs with the max occurrence
 > # increment
 × # repeat each unique letter that many times
 .; # replace the first occurrence of the letter in the input with this

share|improve this answer

edited 17 mins ago

answered 29 mins ago

Emigna

43.5k431131

share|improve this answer

share|improve this answer

edited 17 mins ago

edited 17 mins ago

edited 17 mins ago

answered 29 mins ago

Emigna

43.5k431131

answered 29 mins ago

Emigna

43.5k431131

answered 29 mins ago

Emigna

43.5k431131

43.5k431131

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The input is meant to be able to contain mixed case, then be treated case insensitively e.g. Pip -> Ppii; so this does not currently fit spec. However as Jo king just asked we do not know if ppii or PPII would be acceptable...
  – Jonathan Allan
  5 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @JonathanAllan: Yeah, I'm following the comments to see how the specs are finalized.
  – Emigna
  1 min ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The input is meant to be able to contain mixed case, then be treated case insensitively e.g. Pip -> Ppii; so this does not currently fit spec. However as Jo king just asked we do not know if ppii or PPII would be acceptable...
  – Jonathan Allan
  5 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @JonathanAllan: Yeah, I'm following the comments to see how the specs are finalized.
  – Emigna
  1 min ago
  

  
  
  
  

The input is meant to be able to contain mixed case, then be treated case insensitively e.g. Pip -> Ppii; so this does not currently fit spec. However as Jo king just asked we do not know if ppii or PPII would be acceptable...
– Jonathan Allan
5 mins ago

The input is meant to be able to contain mixed case, then be treated case insensitively e.g. Pip -> Ppii; so this does not currently fit spec. However as Jo king just asked we do not know if ppii or PPII would be acceptable...
– Jonathan Allan
5 mins ago

@JonathanAllan: Yeah, I'm following the comments to see how the specs are finalized.
– Emigna
1 min ago

@JonathanAllan: Yeah, I'm following the comments to see how the specs are finalized.
– Emigna
1 min ago

add a comment | 

up vote
2
down vote

JavaScript (Node.js), 126 bytes

f=x=>[...x].map(c=>(f[c]=-~f[c],f[c]>w?w=f[c]:w,c),w=0).map(c=>x=x.replace(c,c.repeat(parseInt(c,36)>9?w-f[c]+1:1),f[c]=w))&&x

Try it online!

all lower or ALL UPPER only.

Explanation

f = x => // The function. `f=` is needed because we will
 // use the function object as temporary storage
 [...x].map(c => ( // - Iterate among each character...
 f[c] = -~f[c], // - Add one to the character counter
 f[c] > w ? w = f[c] : w, // - Update the maximum count if necessary
 c // - Return back the character for the use in
 // the next map function
 ), w = 0) // - The max counter
 .map(c => // - Iterate again...
 x = x.replace( // - Repeat the first appearance of
 c, // - Each character
 c.repeat( // - Needed number times
 parseInt(c, 36) > 9 // - Additional constraint
 ? w - f[c] + 1 // - If this is letter, repeat
 : 1 // - If not, stay as is
 ), // - That should've been clearly stated
 f[c] = w // - And set the counter so that no further 
 // replacements are done on this character 
 ) // - (w - f[c] + 1 = 1 in further iterations)
 ) && x // - Return the result

share|improve this answer

edited 45 mins ago

answered 4 hours ago

Shieru Asakoto

1,690311

add a comment | 

up vote
2
down vote

JavaScript (Node.js), 126 bytes

f=x=>[...x].map(c=>(f[c]=-~f[c],f[c]>w?w=f[c]:w,c),w=0).map(c=>x=x.replace(c,c.repeat(parseInt(c,36)>9?w-f[c]+1:1),f[c]=w))&&x

Try it online!

all lower or ALL UPPER only.

Explanation

f = x => // The function. `f=` is needed because we will
 // use the function object as temporary storage
 [...x].map(c => ( // - Iterate among each character...
 f[c] = -~f[c], // - Add one to the character counter
 f[c] > w ? w = f[c] : w, // - Update the maximum count if necessary
 c // - Return back the character for the use in
 // the next map function
 ), w = 0) // - The max counter
 .map(c => // - Iterate again...
 x = x.replace( // - Repeat the first appearance of
 c, // - Each character
 c.repeat( // - Needed number times
 parseInt(c, 36) > 9 // - Additional constraint
 ? w - f[c] + 1 // - If this is letter, repeat
 : 1 // - If not, stay as is
 ), // - That should've been clearly stated
 f[c] = w // - And set the counter so that no further 
 // replacements are done on this character 
 ) // - (w - f[c] + 1 = 1 in further iterations)
 ) && x // - Return the result

share|improve this answer

edited 45 mins ago

answered 4 hours ago

Shieru Asakoto

1,690311

add a comment | 

up vote
2
down vote

up vote
2
down vote

JavaScript (Node.js), 126 bytes

f=x=>[...x].map(c=>(f[c]=-~f[c],f[c]>w?w=f[c]:w,c),w=0).map(c=>x=x.replace(c,c.repeat(parseInt(c,36)>9?w-f[c]+1:1),f[c]=w))&&x

Try it online!

all lower or ALL UPPER only.

Explanation

f = x => // The function. `f=` is needed because we will
 // use the function object as temporary storage
 [...x].map(c => ( // - Iterate among each character...
 f[c] = -~f[c], // - Add one to the character counter
 f[c] > w ? w = f[c] : w, // - Update the maximum count if necessary
 c // - Return back the character for the use in
 // the next map function
 ), w = 0) // - The max counter
 .map(c => // - Iterate again...
 x = x.replace( // - Repeat the first appearance of
 c, // - Each character
 c.repeat( // - Needed number times
 parseInt(c, 36) > 9 // - Additional constraint
 ? w - f[c] + 1 // - If this is letter, repeat
 : 1 // - If not, stay as is
 ), // - That should've been clearly stated
 f[c] = w // - And set the counter so that no further 
 // replacements are done on this character 
 ) // - (w - f[c] + 1 = 1 in further iterations)
 ) && x // - Return the result

share|improve this answer

edited 45 mins ago

answered 4 hours ago

Shieru Asakoto

1,690311

JavaScript (Node.js), 126 bytes

f=x=>[...x].map(c=>(f[c]=-~f[c],f[c]>w?w=f[c]:w,c),w=0).map(c=>x=x.replace(c,c.repeat(parseInt(c,36)>9?w-f[c]+1:1),f[c]=w))&&x

Try it online!

all lower or ALL UPPER only.

Explanation

f = x => // The function. `f=` is needed because we will
 // use the function object as temporary storage
 [...x].map(c => ( // - Iterate among each character...
 f[c] = -~f[c], // - Add one to the character counter
 f[c] > w ? w = f[c] : w, // - Update the maximum count if necessary
 c // - Return back the character for the use in
 // the next map function
 ), w = 0) // - The max counter
 .map(c => // - Iterate again...
 x = x.replace( // - Repeat the first appearance of
 c, // - Each character
 c.repeat( // - Needed number times
 parseInt(c, 36) > 9 // - Additional constraint
 ? w - f[c] + 1 // - If this is letter, repeat
 : 1 // - If not, stay as is
 ), // - That should've been clearly stated
 f[c] = w // - And set the counter so that no further 
 // replacements are done on this character 
 ) // - (w - f[c] + 1 = 1 in further iterations)
 ) && x // - Return the result

share|improve this answer

edited 45 mins ago

answered 4 hours ago

Shieru Asakoto

1,690311

share|improve this answer

share|improve this answer

edited 45 mins ago

edited 45 mins ago

edited 45 mins ago

answered 4 hours ago

Shieru Asakoto

1,690311

answered 4 hours ago

Shieru Asakoto

1,690311

answered 4 hours ago

Shieru Asakoto

1,690311

1,690311

add a comment | 

add a comment | 

up vote
2
down vote

J, 44 bytes

#~1+(</"1@=*a)+/@(*>./-])#/.~*a=.1='`z'I.~.

Try it online!

share|improve this answer

edited 15 mins ago

answered 3 hours ago

FrownyFrog

2,1671518

add a comment | 

up vote
2
down vote

J, 44 bytes

#~1+(</"1@=*a)+/@(*>./-])#/.~*a=.1='`z'I.~.

Try it online!

share|improve this answer

edited 15 mins ago

answered 3 hours ago

FrownyFrog

2,1671518

add a comment | 

up vote
2
down vote

up vote
2
down vote

J, 44 bytes

#~1+(</"1@=*a)+/@(*>./-])#/.~*a=.1='`z'I.~.

Try it online!

share|improve this answer

edited 15 mins ago

answered 3 hours ago

FrownyFrog

2,1671518

J, 44 bytes

#~1+(</"1@=*a)+/@(*>./-])#/.~*a=.1='`z'I.~.

Try it online!

share|improve this answer

edited 15 mins ago

answered 3 hours ago

FrownyFrog

2,1671518

share|improve this answer

share|improve this answer

edited 15 mins ago

edited 15 mins ago

edited 15 mins ago

answered 3 hours ago

FrownyFrog

2,1671518

answered 3 hours ago

FrownyFrog

2,1671518

answered 3 hours ago

FrownyFrog

2,1671518

2,1671518

add a comment | 

add a comment | 

up vote
1
down vote

Perl 6, 79 bytes

->aa.=subst($_,$_ x a.comb(/<:Ll>/).Bag.values.max+1-a.comb($_))for 'a'..'z'

Try it online!

Takes a mutable string and edits it in place. This makes no distinction between case, so input should be either all upper or all lower case. Oof, this feels like it could be shorter.

Explanation:

->a # Anonymous code block that takes a mutable string 
 for 'a'..'z' # For each letter
 a.=subst( ) # Substitute
 $_, #The first occurrence of the letter with
 $_ x #The letter repeated
 a.comb(/<:Ll>/) The lowercase letters of the string
 .Bag.values.max # The count of the most common letter
 +1 # Plus 1
 -a.comb($_) # Minus the count of that letter already in the string

share|improve this answer

edited 19 mins ago

answered 4 hours ago

Jo King

16.8k24193

add a comment | 

up vote
1
down vote

Perl 6, 79 bytes

->aa.=subst($_,$_ x a.comb(/<:Ll>/).Bag.values.max+1-a.comb($_))for 'a'..'z'

Try it online!

Takes a mutable string and edits it in place. This makes no distinction between case, so input should be either all upper or all lower case. Oof, this feels like it could be shorter.

Explanation:

->a # Anonymous code block that takes a mutable string 
 for 'a'..'z' # For each letter
 a.=subst( ) # Substitute
 $_, #The first occurrence of the letter with
 $_ x #The letter repeated
 a.comb(/<:Ll>/) The lowercase letters of the string
 .Bag.values.max # The count of the most common letter
 +1 # Plus 1
 -a.comb($_) # Minus the count of that letter already in the string

share|improve this answer

edited 19 mins ago

answered 4 hours ago

Jo King

16.8k24193

add a comment | 

up vote
1
down vote

up vote
1
down vote

Perl 6, 79 bytes

->aa.=subst($_,$_ x a.comb(/<:Ll>/).Bag.values.max+1-a.comb($_))for 'a'..'z'

Try it online!

Takes a mutable string and edits it in place. This makes no distinction between case, so input should be either all upper or all lower case. Oof, this feels like it could be shorter.

Explanation:

->a # Anonymous code block that takes a mutable string 
 for 'a'..'z' # For each letter
 a.=subst( ) # Substitute
 $_, #The first occurrence of the letter with
 $_ x #The letter repeated
 a.comb(/<:Ll>/) The lowercase letters of the string
 .Bag.values.max # The count of the most common letter
 +1 # Plus 1
 -a.comb($_) # Minus the count of that letter already in the string

share|improve this answer

edited 19 mins ago

answered 4 hours ago

Jo King

16.8k24193

Perl 6, 79 bytes

->aa.=subst($_,$_ x a.comb(/<:Ll>/).Bag.values.max+1-a.comb($_))for 'a'..'z'

Try it online!

Takes a mutable string and edits it in place. This makes no distinction between case, so input should be either all upper or all lower case. Oof, this feels like it could be shorter.

Explanation:

->a # Anonymous code block that takes a mutable string 
 for 'a'..'z' # For each letter
 a.=subst( ) # Substitute
 $_, #The first occurrence of the letter with
 $_ x #The letter repeated
 a.comb(/<:Ll>/) The lowercase letters of the string
 .Bag.values.max # The count of the most common letter
 +1 # Plus 1
 -a.comb($_) # Minus the count of that letter already in the string

share|improve this answer

edited 19 mins ago

answered 4 hours ago

Jo King

16.8k24193

share|improve this answer

share|improve this answer

edited 19 mins ago

edited 19 mins ago

edited 19 mins ago

answered 4 hours ago

Jo King

16.8k24193

answered 4 hours ago

Jo King

16.8k24193

answered 4 hours ago

Jo King

16.8k24193

16.8k24193

add a comment | 

add a comment | 

 

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