Given a string s and a positive integer N, gradually duplicate each character more and more until N duplicates, and then staying at N duplicates until N characters away from the end, then step down again.

For example, given abalone and 3:

a    we start with 1 copy
bb   then 2 copies
aaa  then 3 copies, which is our second parameter
lll  so we continue using 3 copies
ooo  until we reach the end
nn   where we use 2 copies
e    and then finally 1 copy

and the result would be abbaaalllooonne.

It is guaranteed that the string has length greater than 2N and only has characters from a to z.

More testcases:

N string       output
2 aaaaa        aaaaaaaa
3 abcdabcdabcd abbcccdddaaabbbcccdddaaabbbccd

This is . Shortest answer in bytes wins. Standard loopholes apply.


17 Answers 17


Jelly, 6 bytes


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How it works

JṡFṢị⁸  Main link. Arguments: s (string), n (integer)

J       Get the indices of s.
 ṡ      Split the indices into overlapping chunks of length n.
  F     Flatten the array of chunks.
   Ṣ    Sort the resulting array of indices.
    ị⁸   Get the characters of s at these indices.

Sample run

JṡFṢị⁸  "abalone", 3

J       [1, 2, 3, 4, 5, 6, 7].
 ṡ      [[1, 2, 3], [2, 3, 4], [3, 4, 5], [4, 5, 6], [5, 6, 7]]
  F     [1, 2, 3, 2, 3, 4, 3, 4, 5, 4, 5, 6, 5, 6, 7]
   Ṣ    [1, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5, 6, 6, 7]
    ị⁸  "abbaaalllooonne"
  • 3
    \$\begingroup\$ That split+flatten+sort method is pure genius. Nice! :) \$\endgroup\$ – HyperNeutrino Jul 4 '17 at 19:04

Python 2, 57 bytes

f=lambda s,n,i=1:s and s[0]*len(s[:i][:n])+f(s[1:],n,i+1)

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Also 57:

Python 2, 57 bytes

f=lambda s,n,i=1:s and s[0]*len(s[:i])+f(s[1:],n,i+(i<n))

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  • \$\begingroup\$ Can you explain your logic behind len(s[:i][:n])? I'm convinced there's a shorter way to get that number but I'm not sure how. \$\endgroup\$ – musicman523 Jul 4 '17 at 6:49
  • \$\begingroup\$ Never mind, I got it! But it's one byte shorter than min(len(s),i,n). Great job! \$\endgroup\$ – musicman523 Jul 4 '17 at 6:53

JavaScript (ES6), 67 65 bytes

-2 bytes thanks to Chas Brown's shorter method using min().


Takes input in currying syntax: f("abalone")(3).

Test Snippet

<div oninput="O.value=f(S.value)(+N.value)">String: <input id=S> N: <input id=N size=3></div>Out: <input id=O size=50 disabled>


Jelly, 8 7 bytes


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How it Works

J««U$⁸x - main link, input e.g. abalone
J        - range of length of letters: [1,2,3,4,5,6,7]
 «       - minimum of each term with second input: [1,2,3,3,3,3,3]
  «U$    - termwise minimum with the reverse: 
     ⁸x  - repeat each character of the input a number of times corresponding to elements:
                    a*1;b*2;a*3...e*1 = abbaaalllooonne

-1 byte thanks to @LeakyNun

  • \$\begingroup\$ Good find @LeakyNun! The closest I got in that direction was J«¥@«U$x@ for 9 bytes. \$\endgroup\$ – fireflame241 Jul 4 '17 at 5:18
  • \$\begingroup\$ Explanation please? \$\endgroup\$ – Comrade SparklePony Jul 4 '17 at 5:20
  • \$\begingroup\$ @fireflame241 genenrally, x@⁸ is equivallent to ⁸x (I used here) \$\endgroup\$ – Leaky Nun Jul 4 '17 at 5:38

Haskell, 61 60 bytes

Thanks to @Laikoni for helping to shave off 1 byte

n#s=do(i,c)<-zip[1..]s;replicate(minimum[n,i,length s-i+1])c

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(#) n string = do
    (i, char) <- zip [1..] string
    replicate (minimum [n, i, length(string)-i+1]) char
  • \$\begingroup\$ Great use of a do block! Save a byte by dropping the parenthesis in length(s). \$\endgroup\$ – Laikoni Jul 5 '17 at 5:28

Haskell (Lambdabot), 74 bytes

f x n=join$zipWith r([1..n]++r(length x-2*n)n++reverse[1..n])x

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  • \$\begingroup\$ Imports count in the score! You would be better of with >>=id \$\endgroup\$ – bartavelle Jul 4 '17 at 6:57
  • \$\begingroup\$ Yeah, I did that too before and then I saw this (that's why there's Lambdabot in brackets). What is the right way? \$\endgroup\$ – ბიმო Jul 4 '17 at 7:54
  • \$\begingroup\$ I stand corrected, I think this is alright! \$\endgroup\$ – bartavelle Jul 4 '17 at 8:01
  • \$\begingroup\$ Good to know, there are a lot of very handy imports in that list. \$\endgroup\$ – ბიმო Jul 4 '17 at 8:03

J, 24 bytes


The bit in parens -- (<.&n<./(|.,:[)>:i.#s) -- creates the 1 2 ... n n n ... 2 1 array, as follows:

                   #s    length of s, call it L
                 i.      numbers 0 1 ... L-1
               >:        increment by 1, now 1 2 ... L
        (|.,:[)          fork: |. = reverse, ,: = stack, [ = identity
                         resulting in  L ... 2 1
                                       1 2 ... L 
     <./                 min of each element of the top and bottom row
 <.&n                    min of each resulting elm and n

once we have that, J's # operator automatically does exactly what asked for, duplicating each element the number of times specified.

Curious to see a J expert's improvment on this...

  • \$\begingroup\$ 23 bytes with a quite different approach [#~#@[$([:>:<:,&:i.-)@] (maybe a stray space got caught in there). I'm at a loss as to why the hook isn't taking x but not in too much a position to care. \$\endgroup\$ – cole Sep 5 '17 at 0:36

PHP>=7.1, 75 bytes


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PHP>=7.1, 78 bytes


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PHP>=7.1, 80 bytes

for([,$a,$n]=$argv;$i<$l=strlen($a);)echo str_repeat($a[$i],min($n,$l-$i,++$i));

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Japt, 11 10 bytes


Test it


Implicit input of string U and integer V.


Map over U and replace every character.


Get the minimum of V, ...


T (initially 0) incremented by 1, ...


And the index of the current character (E) subtracted from (n) the length (Ê) of U.


Repeat the current character that many times.

Implicitly output the final string.


R, 87 bytes


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Python 2 68 bytes

f=lambda s,n:''.join(s[i]*min(i+1,len(s)-i,n)for i in range(len(s)))
  • \$\begingroup\$ You don't need the f= in the answer; the function can be anonymous. With that in mind, you can remove 3 bytes with lambda s,n:''.join(c*min(i+1,len(s)-i,n)for i,c in enumerate(s)). \$\endgroup\$ – notjagan Jul 4 '17 at 4:44

Husk, 10 9 bytes


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The first line is the main function, it repeats each letter n times and then calls the second line twice.

The second line takes at most N letters from each group of repeated letters, where N is the 1-based index of the group, then reverses the list.


Haskell, 68 bytes

g 1
g _""_=""
g i(a:b)n=(a<$[1..min(length b+1)$min i n])++g(i+1)b n

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APL (Dyalog), 15 bytes


{} function where left argument (cap) is and right argument (string) is :

≢⍵ count the number of characters in the string

 generate that many ɩntegers

i← store in i


i⌊ pairwise minimum with i

⍺⌊ pairwise minimum with the cap

⍵/⍨ use those numbers to replicate the letters of the string

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F#, 96 bytes

let f(s:string)n=Seq.mapi(fun i c->System.String(c,Seq.min[i+1;s.Length-i;n]))s|>String.concat""

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A port of Justin Mariners javascript answer


Java (OpenJDK 8), 101 97 bytes

n->s->{int i=0,j;for(char c:s)for(j=s.length-i++,j=j<i?j:i,j=j<n?j:n;j-->0;)System.out.print(c);}

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So much playing with indices...


Abalone is a type of fish (well, a shellfish), therefore…

><>, 79 bytes


Try it online, or watch it at the fish playground!

Reads the string from STDIN, and assumes the number is already on the stack.

Explanation: The second, fourth and sixth lines are the main loops. The details are some ugly stack manipulation, but in broad strokes, first, the second line fills the stack alternating between a character of input and min(in), where n is the length cap and i is the index of the character in the input: for "abalone", 3, the stack looks like

"a", 1, "b", 2, "a", 3, "l", 3, "o", 3, "n", 3, "e", 3, -1=EOF, 3

Next, line 4 goes through the stack in the same way in reverse, to get the right hand end capped properly:

"a", 1, "b", 2, "a", 3, "l", 3, "o", 3, "n", 2, "e", 1, -1

Then the sixth line takes each character–number pair and prints the character as many times as the number.


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