Write a program to elasticize strings

Question

Nice verb there, in the title.

Write a program that given an input string, will "elasticize" this string and output the result. Elasticizing a string is done as follows:

The first character is shown once. The second character is shown twice. The third character is shown thrice, and so on.

As you can see, the amount of duplications of a certain character is related to the character's index as opposed to its previous occurrences in the string.

You can expect to receive only printable ASCII characters. Based off the following link, these characters have decimal values 32-126.

Examples:

Why: Whhyyy

SKype: SKKyyyppppeeeee

LobbY: LoobbbbbbbYYYYY (Note how there are 7 b's since the first b is shown 3 times and the second b is shown 4 times, making a total of 7 b's).

A and B: A aaannnnddddd BBBBBBB

Shortest bytes wins :)

Answer 1

Jelly, 3 bytes

Code:

ĖP€

Explanation:

Ė     # Enumerate.
 P€   # Product of each.
      # Implicit joining of everything.

Uses the Jelly encoding. Try it online!.

Answer 2

J, 4 bytes

#~#\

Usage

   f =: #~#\
   f 'Why'
Whhyyy
   f 'SKype'
SKKyyyppppeeeee
   f 'LobbY'
LoobbbbbbbYYYYY
   f 'A and B'
A  aaannnnddddd      BBBBBBB

Explanation

#~#\  Input: s
  #\  Computes the length of each prefix of s
      This forms the range [1, 2, ..., len(s)]
#~    For each value in the range, copy the character at the
      corresponding index that many times
      Return the created string

Answer 3

Brainfuck, 15 bytes

,[>+[->+<<.>],]

Pretty straightforward implementation, shifting the memory space by 1 for each input char. Requires an interpreter that gives 0 on EOF, and 32-bit/arbitrary precision cells for inputs longer than 255 chars.

Try it online! (Note: TIO uses 8-bit cells)

Answer 4

Haskell, 29 bytes

concat.zipWith replicate[1..]

Usage example: concat.zipWith replicate[1..] $ "SKype" -> "SKKyyyppppeeeee".

replicate n c makes n copies of c and concat makes a single list out of all the sublists.

Answer 5

Java, 158 121 bytes

Saved a whopping 37 bytes thanks to Kevin Cruijssen!

interface a{static void main(String[]A){int b=0,B;for(char c:A[0].toCharArray())for(B=b+++2;--B>0;)System.out.print(c);}}

As a bonus, this program can handle all Unicode characters in the existence, including the control characters located at the very end of Basic Multilingual Plane.

Answer 6

Perl, 16 bytes

s/./$&x$+[0]/ge

+1 byte for the -p flag.

s/./        /    find every character
             g   globally
              e  and replace with the eval'd result of
    $&           the matched string
      x          repeated
       $+[0]     by the index of the character after the match

Answer 7

Python, 39 bytes

f=lambda s:s and f(s[:-1])+s[-1]*len(s)

Test it on Ideone.

Answer 8

CJam, 9 8 7 bytes

Thanks to jimmy23013 for saving 1 byte.

Sl+eee~

Test it here.

Explanation

Using the LobbY example:

                                      Stack:
S    e# Push space.                   [" "]
l    e# Read input.                   [" " "LobbY"]
+    e# Append.                       [" LobbY"]
ee   e# Enumerate.                    [[[0 ' ] [1 'L] [2 'o] [3 'b] [4 'b] [5 'Y]]]
e~   e# Run-length decode.            ["LoobbbbbbbYYYYY"]

Answer 9

MATLAB, 45 bytes

g=@(m)sort(m(m>0));@(s)s(g(hankel(1:nnz(s))))

Explanation: The key is hankel, which produces a Hankel matrix of a given vector. From this matrix, we can extract a vector of indices, which defines which character of the string is at which position in the output. E.g. hankel(1:4) produces following matrix:

 1  2  3  4
 2  3  4  0
 3  4  0  0
 4  0  0  0

From this matrix we can extrac the vector 1,2,2,3,3,3,4,4,4,4,4. This vector allows us to output the first character of the string once, the second one twice e.t.c.

Answer 10

Javascript ES6, 39 bytes

x=>x.replace(/./g,(y,i)=>y+y.repeat(i))

Same length, but more fun:

x=>x.replace(i=/./g,y=>y.repeat(i=-~i))

Snippet demo:

f= x=>x.replace(/./g,(y,i)=>y+y.repeat(i))
run.onclick=_=>output.textContent=f(input.value)

<input id="input" value="SKype">
<button id="run">Go</button>
<pre id="output"></pre>

Answer 11

APL (dzaima/APL), 6 5 bytes

⍳∘≢⌿⊢

Try it online!

⍳∘≢ enumeration of the argument... (indices of its length)
⌿ replicates the elements of...
⊢ the unmodified argument

Answer 12

APL (8)

{⍵/⍨⍳⍴⍵}

I.e.:

      {⍵/⍨⍳⍴⍵} ¨  'Why' 'SKype' 'LobbY'
┌──────┬───────────────┬───────────────┐
│Whhyyy│SKKyyyppppeeeee│LoobbbbbbbYYYYY│
└──────┴───────────────┴───────────────┘

Explanation:

• ⍴⍵: length of given vector
• ⍳: numbers 1..N
• ⍵/⍨: replicate each element in ⍵ N times.

Answer 13

Brachylog, 13 bytes

:ImC,0:Ie,Cw\

This prints the result to STDOUT.

Explanation

This is a good example of exploiting backtracking to loop.

:ImC            C is the Ith character of the Input
    ,
     0:Ie       Unify an implicit variable with an integer between 0 and I
         ,
          Cw    Write C to STDOUT
            \   False, trigger backtracking. It will go back to 0:Ie and unify the implicit
                variable with another integer, until all integers were used. After that, it
                will backtrack to :ImC and unify I and C with the next character.

Answer 14

Javascript ES6, 42 41 bytes

s=>[,...s].map((e,i)=>e.repeat(i)).join``

Example runs:

f=s=>[,...s].map((e,i)=>e.repeat(i)).join``

f("Why")   => "Whhyyy"
f("SKype") => "SKKyyyppppeeeee"
f("LobbY") => "LoobbbbbbbYYYYY"

Answer 15

R, 83 50 bytes

-23 Thanks to Giuseppe, though he used essentially an entire new method altogether

function(s)intToUtf8(rep(utf8ToInt(s),1:nchar(s)))

My original post:

function(s){r="";for(i in 1:nchar(s))r=paste0(r,strrep(el(strsplit(s,""))[i],i));r}

Try it online!

I feel like there's definitely a better way to do this, but with my new knowledge of a few functions in R, this is my approach.

Answer 16

PowerShell v2+, 36 bytes

-join([char[]]$args[0]|%{"$_"*++$i})

Takes input $args[0], explicitly casts it as a char array, sends that into a loop |%{...}. Each iteration we take the current letter/character "$_" and use the * overloaded operator to concatenate the string pre-incremented $i times. The result of each loop iteration is encapsulated in parens to form an array and then -joined together to form a string. That string is left on the pipeline and output is implicit.

Examples

PS C:\Tools\Scripts\golfing> .\elasticize-a-word.ps1 Why
Whhyyy

PS C:\Tools\Scripts\golfing> .\elasticize-a-word.ps1 SKype
SKKyyyppppeeeee

PS C:\Tools\Scripts\golfing> .\elasticize-a-word.ps1 LobbY
LoobbbbbbbYYYYY

PS C:\Tools\Scripts\golfing> .\elasticize-a-word.ps1 'a b'
a  bbb

Answer 17

MATLAB, 23 bytes

@(x)repelem(x,1:nnz(x))

Creates an anonymous function ans that can be called using ans('stringtoelacticize')

Answer 18

K/Kona, 14 bytes

{,/(1+!#x)#'x}

Usage:

k){,/(1+!#x)#'x}"A and B"
"A  aaannnnddddd      BBBBBBB"

Answer 19

Perl 6,  22 20  19 bytes

{S:g/(.)/{$0 x$/.to}/}
{S:g[(.)]=$0 x$/.to}
{[~] .comb Zx 1..*}

Explanation:

{          # implicit parameter $_
  [~]      # string concatenate the following list
    .comb  # the NFG characters from $_
    Z[x]   # zip combined using the string repetition operator
    1 .. * # 1 to infinity
}

Answer 20

VBA, 75 bytes

Function e(s):For a=1 To Len(s):e=e &String(a,Mid(s,a,1)):Next:End Function

Call as e.g. a user function in a spreadsheet.

=e(A1)

┌─────────┬───────────────┐
│   SKype │SKKyyyppppeeeee│
└─────────┴───────────────┘

It truncates if you feed it its own output a few times :-).

Answer 21

Julia, 32 bytes

!s=join(split(s[k=1:end],"").^k)

Unlike Dennis's solution, this is not recursive. split with argument "" separates the string into an array of strings of length 1. The [k=1:end] is a trick to create a range from 1 to the number of characters in the string, and this range is used to concatenate n copies of the n-th character. join then recombines the array of strings into a single string, in order.

Usage example: !"SKype"

Answer 22

PHP, 68 bytes

<?php foreach(str_split($argv[1])as$i=>$a)echo str_repeat($a,$i+1);

Answer 23

Retina, 22 bytes

Byte count assumes ISO 8859-1 encoding.

.
$&$.`$*·
+`(.)·
$1$1

Try it online!

Basically, we insert the right amount of · as placeholders between the characters (since these extended ASCII characters can't appear in the input), then fill them up with the adjacent character in the second stage.

Answer 24

05AB1E, 2 bytes

ƶJ

Try it online or verify all test cases.

Explanation:

ƶ   # Multiply each character in the (implicit) input-string by its 1-based index
 J  # Join the list of substrings together to a single string
    # (after which the result is output implicitly)

Answer 25

Vyxal s, 2 bytes

Þż

Try it online, or see a 3 byte flagless version

Answer 26

BQN, 9 8 bytes

⊢/˜1+↕∘≠

Anonymous tacit function. Try it at BQN online!

Explanation

⊢/˜1+↕∘≠
       ≠  Length of argument
     ↕∘   Range
   1+     Add 1 to each, making it an inclusive range from 1 to the length
 /˜       Replicate that many times the elements of
⊢         The original input

Answer 27

><> (Fish), 33 bytes

1:i:0(?^$:?v~~1+00.
.!02o:$;!-1<|

Try the animated version

Basically if the counter is 0 skip to the second :, otherwise the first :. Properly adjust the counters.

Answer 28

Actually, 7 bytes

' +ñ♂πΣ

Try it online!

Explanation:

' +ñ♂πΣ
' +      prepend a space
   ñ     enumerate ("abc" -> [[0, 'a'], [1, 'b'], [2, 'c']])
    ♂π   map: for each character, repeat it n times
      Σ  concatenate

Answer 29

Pyth - 5 bytes

1 byte saved thanks to @FryAmTheEggman.

s*VSl

Test Suite.

Answer 30

Python 3, 48 47 bytes

Thanks to mego for saving a byte with the -~i trick.

lambda s:''.join(c*-~i for i,c in enumerate(s))

This is mostly self-explanatory. One thing for those not versed in Python: The * operator is overloaded to act like Perl's x operator, repeating its string argument the number of times specified by its numeric argument. E.g. 'foo' * 3 == 'foofoofoo'