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

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 5

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 6

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 7

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 8

Python, 39 bytes

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

Test it on Ideone.

Answer 9

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 10

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 11

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 12

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 13

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 14

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 15

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 16

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 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

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 22

PHP, 68 bytes

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

Answer 23

Pyth - 5 bytes

1 byte saved thanks to @FryAmTheEggman.

s*VSl

Test Suite.

Answer 24

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 25

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'

Answer 26

C#, 81 Bytes

void f(string s){for(int i=0;i<s.Length;i++)Console.Write(new String(s[i],i+1));}

Answer 27

MATL, 5 bytes

tn:Y"

Try it Online

Explanation

    % Implictly grab input as a string
tn  % Duplicate and compute the length (N)
:   % Create an array from [1...N]
Y"  % Perform run-length decoding to elacticize the string
    % Implicitly display the result

Answer 28

Python, 40 bytes

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

Answer 29

Julia, 34 bytes

!s=s>""?!s[1:(e=end)-1]*s[e:e]^e:s

Try it online!

Answer 30

TSQL, 97 bytes

Golfed:

DECLARE @x varchar(max)='Lobby'
DECLARE @ int=LEN(@x)WHILE @>0SELECT
@x=STUFF(@x,@,1,REPLICATE(SUBSTRING(@x,@,1),@)),@-=1PRINT @x

Ungolfed:

DECLARE @x varchar(max)='Lobby'

DECLARE @ int=LEN(@x)
WHILE @>0
  SELECT 
    @x=STUFF(@x,@,1,REPLICATE(SUBSTRING(@x,@,1),@)),
    @-=1

PRINT @x

Try it online

Answer 31

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 32

Mathematica, 46 bytes

""<>Table@@@(#^Range@Length@#&@Characters[#])&

Unnamed function. Takes the characters of the input string and raises them to the power of their position in the string (i.e. "abc" becomes {"a"^1, "b"^2, "c"^3}). Yes, MMA don't give a shit 'bout types :-)

The FullForm of the above list elements is Power["a", 1], etc.
Table@@@ acts on the list, replacing the head of each element (in this case Power) with Table.
Results in {Table["a"], Table["b", 2], Table["c", 3]} (because "a"^1 -> "a").
This evaluates to {"a", {"b", "b"}, {"c", "c", "c"}} Finally the infix concatenation operator <> concatenates this with the empty string.

I exploited the fact that Power is Listable, i.e. it automatically threads over corresponding elements of lists, but not orderless (because "a" + 1 would evaluate to Plus[1, "a"]).

Answer 33

LINQPad w/ C#, 82 bytes

void m(string s){Console.Write(s.SelectMany((x,i)=>new string(x,++i)).ToArray());}

Single output operation.

Answer 34

C, 84 bytes

main(c,v,i,j)char**v;{for(i=0;i<strlen(v[1]);++i)for(j=0;j<=i;++j)putchar(v[1][i]);}

This should compile on gcc with no flags. Input is taken through the first command-line argument. E.g.

$ ./elasticize Why
Whhyyy
$ ./elasticize SKype
SKKyyyppppeeeee
$ ./elasticize LobbY
LoobbbbbbbYYYYY
$ ./elasticize A and B
A  aaannnnddddd      BBBBBBB

Ungolfed:

int main(int argc, char** argv) {
    int i, j;
    for(i = 0; i < strlen(argv[1]); ++i) {
        for(j = 0; j <= i; ++j) {
            putchar(v[1][i]);
        }
    }
}

Answer 35

Clojure, 86 bytes

(fn[s](apply str(mapcat #(repeat(+(nth % 1)1)(nth % 0))(map list s(range(count s))))))

Well, nearly beat C. Create a list of pairs (symbol, its position), then repeat each symbol given number of times and flatten the result list and concatenate the list into one string.

See it here: https://ideone.com/uQNar2

Answer 36

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 37

05AB1E, 6 bytes

vN>Fy?

Explained

v       # for each char in string
 N>F    # index+1 number of times do
    y?  # print current char

Try it online

Answer 38

Retina, 62 bytes

It wasn't as easy or short as I thought it'd be. Note that the code contains no spaces. They are all tabs (which are rendered incorrectly here), and the last line is blank.

.*
$0¶ ¶
{+`^(.)(.*)¶    (.*¶.*)
$1$2    ¶$3$1
(   +)¶
¶   $1
}`^.

    |¶

Try it online

Answer 39

Pyke, 4 bytes

Foh*

Try it here!

Answer 40

Python 2.7 - 47 Bytes

''.join([s[i-1]*i for i in range(1, len(s)+1)])

where 's' is the given string Output:

welcome: weelllccccooooommmmmmeeeeeee
00004:   000000000044444
Why:     Whhyyy
SKype:   SKKyyyppppeeeee
A and B: A  aaannnnddddd      BBBBBBB

Answer 41

Oracle SQL 11.2, 125 bytes

SELECT LISTAGG(SUBSTR(RPAD(' ',LEVEL+1,SUBSTR(:1,LEVEL,1)),2))WITHIN GROUP(ORDER BY 1)FROM DUAL CONNECT BY LEVEL<=LENGTH(:1);

Answer 42

C, 77 bytes

Not much room for golfing here. If only there were a string repeat operator.

i;main(j){char s[999];gets(s);for(;s[i];i++)for(j=0;j<=i;j++)putchar(s[i]);}

Try it online! http://ideone.com/UliJfD

Answer 43

Jelly, 2 bytes (non-competing)

xJ

Try it online!

Answer 44

Javascript, 58 bytes

e=a=>(Array.from(a).map((e,i)=>(e).repeat(i+1))).join("");

Answer 45

Java, 136 bytes

enum b{;public static void main(String[]a){int i=0,j=0;for(;i<a[0].length();i++){for(;j<i+1;j++)System.out.print(a[0].charAt(i));j=0;}}}

Takes input as program arguments.

Ungolfed with comments:

enum b {;
    public static void main(String[]a) {
        int i=0, j=0;                               // Init counters
        for (; i<a[0].length(); i++) {              // For each characters in string
            for (; j < i + 1; j++)                  // Loop from 0 to i
                System.out.print(a[0].charAt(i));   // Print character at i
            j=0;                                    // Reset j
        }
    }
}

Answer 46

PHP, 76 bytes

I'm sure this can be improved...

<?php for($i=0;$i<strlen($argv[1]);$i++)echo str_repeat($argv[1]{$i},$i+1);

Run from command line:

$ php [file] "Testing"

Answer 47

Ruby, 29 bytes

Try it online

->s{i=0;s.gsub(/./){$&*i+=1}}

Answer 48

C#, 80 Bytes

x.Select((c,i)=>new{c=c,i=i+1}).ToList().ForEach(o=>Write(new string(o.c,o.i)));

Where x is a string.

required: using static System.Console;

If Microsoft had implemented ForEach on IEnumerable<T>, (which is incredibly easy to do), this would be shorter by 9 bytes by removing the .ToList()

Answer 49

Silicon(non-competing), 6 bytes

(This language is exactly the same age as this challenge, but I added some useful commands right after I saw it, so non-competing I think.)

biSÚÿj

Silicon uses CP037, which is a 255-bit codepage.

Explanation:

b        Push a space
i        Input
S        Split
Ú        Enumerate
ÿ        Replicate each list item n times
j        Join