Multiply a string by a number!

Question

There was a challenge up a while ago about multiplying strings. It showed us how we can multiply not only numbers, but also strings. However, we still can't multiply a number by a string properly. There has been one attempt to do so but this is obviously wrong. We need to fix that!

Your Task:

Write a function or program that multiplies two inputs, a string and an integer. To (properly) multiply an string by an integer, you split the string into characters, repeat each character a number of times equal to the integer, and then stick the characters back together. If the integer is negative, we use its absolute value in the first step, and then reverse the string. If the input is 0, output nothing (anything multiplied by 0 equals nothing).

Input:

A string that consists solely of printable ASCII characters and newlines, and an integer (possible negative).

Output:

The string multiplied by the integer.

Examples:

Hello World!, 3            --> HHHeeellllllooo   WWWooorrrlllddd!!!
foo, 12                    --> ffffffffffffoooooooooooooooooooooooo
String, -3                 --> gggnnniiirrrtttSSS
This is a fun challenge, 0 --> 
Hello
World!, 2                  --> HHeelllloo

                               WWoorrlldd!!

Scoring:

This is code-golf, lowest byte count wins!

Answer 1

Jelly, 6 5 4 bytes

²Ɠxm

Try it online!

How it works

²Ɠxm  Main link. Argument: n (integer)

²     Yield n².
 Ɠ    Read and eval one line of input. This yields a string s.
  x   Repeat the characters of s in-place, each one n² times.
   m  Takes each |n|-th character of the result, starting with the first if n > 0, 
      the last if n < 0.

Answer 2

Python 2, 59 57 50 46 bytes

-2 bytes thanks to Anders Kaseorg. -4 bytes thanks to Dennis.

lambda s,n:''.join(i*n**2for i in s)[::n or 1]

Try it online!

Answer 3

JavaScript (ES6), 63 bytes

Takes input in currying syntax (s)(n).

s=>n=>[...s].reduce((s,c)=>n<0?c.repeat(-n)+s:s+c.repeat(n),'')

Test cases

let f =

s=>n=>[...s].reduce((s,c)=>n<0?c.repeat(-n)+s:s+c.repeat(n),'')

console.log(f(`Hello World!`)(3))
console.log(f(`foo`)(12))
console.log(f(`String`)(-3))
console.log(f(`This is a fun challenge`)(0))
console.log(f(`Hello
World!`)(2))

Answer 4

Python 3, 44 bytes

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

Try it online!

Answer 5

05AB1E, 10 bytes

S²Ä×J²0‹iR

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S          # Split the string into characters
 ²Ä×       # Repeat each character abs(integer) times
    J      # Join into a string
     ²0‹i  # If the integer is less than 0...
         R #   Reverse the string

Answer 6

V, 29, 23, 18, 17 bytes

æ_ñÀuñÓ./&ò
ÀäëÍî

Try it online!

Hexdump:

00000000: e65f f1c0 75f1 d32e 2f26 f20a c0e4 ebcd  ._..u.../&......
00000010: ee                                       .

Thanks to @nmjcman101 for saving 6 bytes, which encouraged me to save another 5!

The original revision was pretty terrible, but now I'm really proud of this answer because it handles negative numbers surprisingly well. (V has next to no numerical support and no support for negative numbers)

Explanation:

æ_          " Reverse the input
  ñ  ñ      " In a macro:
   À        "   Run the arg input. If it's positive it'll give a count. If it's negative
            "   running the '-' will cause V to go up a line which will fail since we're
            "   on the first line, which will break out of this macro
    u       "   (if arg is positive) Undo the last command (un-reverse the line)
      Ó./&ò " Put every character on it's own line

At this point, the buffer looks like this:

H
e
l
l
o

w
o
r
l
d
!
<cursor>

It's important to not the trailing newline, and that the cursor is on it.

À           " Run arg again. If it's negative, we will move up a line, and then give the 
            " absolute value of the count. If it's positive (or 0) it'll just give the
            " count directly (staying on the last line)
 ä          " Duplicate... (count times)
  ë         "   This column. 
   Íî       " Remove all newlines.

Answer 7

Haskell, 41 36 bytes

f n|n<0=reverse.f(-n)|1<3=(<*[1..n])

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Example usage: f (-3) "abc" yields "cccbbbaaa".

Edit: -5 bytes thanks to xnor!

Answer 8

R, 83 78 76 bytes

function(s,i)cat('if'(i<0,rev,`(`)(rep(el(strsplit(s,'')),e=abs(i))),sep='')

Anonymous function.

Frederic saved 3 bytes, Giuseppe saved 2 4.

Explanation:

     el(strsplit(s,''))                      # split string into list characters
 rep(                  ,e=abs(i)))           # repeat each character abs(i) times


    'if'(i<0,rev,   ){...}                 # if i>0, reverse character list
                 `(`                       # otherwise leave it alone: `(` is the identity function
cat(                      ,sep='')         # print the result

Tests:

> f('Hello World!', 3 )
HHHeeellllllooo   WWWooorrrlllddd!!!
> f('foo', 12)
ffffffffffffoooooooooooooooooooooooo
> f('String', -3)
gggnnniiirrrtttSSS
> f('This is a fun challenge', 0)
> f('Hello
+ World!', 2)
HHeelllloo

WWoorrlldd!!

Answer 9

MATL, 9 bytes

y|Y"w0<?P

Inputs are: number, then string.

Strings with newlines are input using char 10 as follows: ['first line' 10 'second line'].

Try it online! Or verify all test cases.

Explanation

Consider inputs -3 and 'String'.

y      % Implicitly take two inputs. Duplicate from below
       % STACK: -3, 'String', -3
|      % Absolute value
       % STACK: -3, 'String', 3
Y"     % Run-length decoding
       % STACK: -3, 'SSStttrrriiinnnggg'
w      % Swap
       % STACK: 'SSStttrrriiinnnggg', -3
0<     % Less than 0?
       % STACK: 'SSStttrrriiinnnggg', 1
?      % If so
  P    %   Flip
       %   STACK: 'gggnnniiirrrtttSSS'
       % End (implicit). Display (implicit)

Answer 10

05AB1E, 10 bytes

0‹FR}ʒ¹Ä×?

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Explanation

0‹F         # input_1 < 0 times do:
   R        # reverse input_2
    }       # end loop
     ʒ      # filter
      ¹Ä×   # repeat current char abs(input_1) times
         ?  # print without newline

Answer 11

PHP>=7.1, 65 bytes

for([,$s,$n]=$argv;$i<strlen($s)*abs($n);)echo$s[$i++/$n-($n<0)];

PHP Sandbox Online

Answer 12

J, 19 15 13 bytes

(#~|)A.~0-@>]

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Explanation

        0-@>]      NB. first or last index depending on sign of right arg
     A.~           NB. get first or last Anagram of left arg
(#~|)              NB. copy left arg, absolute-value-of-right-arg times

Answer 13

Brain-Flak (BrainHack), 154 152 bytes

([(({})(<()>))]<>)<>{({}()<([{}]()<([{}])>)<>({}<>)<>>)<>}{}<>{}<>({}<([][()]){{}({<({}<(({}<>)<>)>())>[()]}<{}{}>)([][()])}{}{}<>>){{}{({}<>)<>}(<>)}{}

Try it online!

Just here to give DJMcMayhem some competition. ;)

Explanation

Here's a modified version of DJMcMayhem's explanation

#Compute the sign and negative absolute value 
([(({})<(())>)]<>)<>{({}()<([{}]()<([{}])>)<>({}<>)<>>)<>}{}<>{}<>

#Keep track of the sign
({}<

    #For each char in the input string:
    ([][()])
    {
        {}

        #Push n copies to the alternate stack
        ({<({}<(({}<>)<>)>())>[()]}<{}{}>)

        #Endwhile
        ([][()])
    }{}{}<>

#Push the sign back on
>)

#If so...
{{}

    #Reverse the whole stack
    {({}<>)<>}

    #And toggle over, ending the loop
    (<>)
}

#Pop the counter off
{}

Answer 14

Dyalog APL, 15 bytes

{⌽⍣(⍵<0)⊢⍺/⍨|⍵}

String as a left argument, number as a right argument.

Try it online!

How?

⍺/⍨ - repeat the string

|⍵ - abs(number) times

⌽⍣ - reverse if

(⍵<0) - the number is below 0

Answer 15

MATLAB, 37 bytes

@(s,n)flip(repelem(s,abs(n)),(n<0)+1)

This defines and anonymous function with inputs s: string and n: number.

Example runs:

>> @(s,n)flip(repelem(s,abs(n)),(n<0)+1)
ans = 
    @(s,n)flip(repelem(s,abs(n)),(n<0)+1)

>> f = ans;

>> f('String', 3)
ans =
SSStttrrriiinnnggg

>> f('String', -3)
ans =
gggnnniiirrrtttSSS

>> f('String', 0)
ans =
   Empty matrix: 1-by-0

Answer 16

Java (OpenJDK 8), 99 98 89 87 85 bytes

s->n->{for(int i=s.length*(n<0?n:-n),r=n<0?0:~i;i++<0;)System.out.print(s[(i+r)/n]);}

Try it online!

• -2 bytes thanks to @Xanderhall
• -2 bytes thanks to @Nevay

Answer 17

Brain-Flak (Haskell), 202 192 bytes

(({})<(([({})]<>)){({}()<([{}])<>({}<>)<>>)<>}{}([{}]<><{}>)([][()]){{}({<({}<(({}<>)<>)>[()])>()}<{}{}>)([][()])}{}{}<>>)([({}<(())>)](<>)){({}())<>}{}{((<{}>))<>{}}{}<>{}{{}{({}<>)<>}(<>)}{}

Try it online!

This is probably the worst possible language to do it in, but it's done. Thanks to @Wheatwizard for providing the Haskell interpreter, which allows mixed input formats. This would be about 150 bytes longer without it.

Explanation:

#Keep track of the first input (n)
(({})<

    #Push abs(n) (thanks WheatWizard!)
    (([({})]<>)){({}()<([{}])<>({}<>)<>>)<>}{}([{}]<><{}>)

    #For each char in the input string:
    ([][()])
    {
        {}

        #Push n copies to the alternate stack
        ({<({}<(({}<>)<>)>[()])>()}<{}{}>)

        #Endwhile
        ([][()])
    }{}{}<>

#Push the original n back on
>)

#Push n >= 0
([({}<(())>)](<>)){({}())<>}{}{((<{}>))<>{}}{}<>{}

#If so...
{{}

    #Reverse the whole stack
    {({}<>)<>}

    #And toggle over, ending the loop
    (<>)
}

#Pop the counter off
{}

Answer 18

Octave, 49 bytes

@(s,n){t=repmat(s,abs(n),1)(:)',flip(t)}{2-(n>0)}

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I will provide an explanation tomorrow.

Answer 19

Pyth, 13 11 bytes

*sm*.aQdz._

Try it!

-2 bytes thanks to @jacoblaw

explanation

*sm*.aQdz._   
  m     z     # map onto the input string (lambda var: d)
   *.aQd      # repeat the char d as often as the absolute value of the input number 
 s            # sum the list of strings into a single string
*        ._Q   # Multiply with the sign of the implicit input value: reverse for negative Q 

old approach, 13 bytes

_W<Q0sm*.aQdz

Try it!

Answer 20

Ruby, 59 +1 = 60 bytes

Uses -n flag.

n=eval$_
a=$<.read
a.reverse!if n<0
a.chars{|i|$><<i*n.abs}

Try it online!

Answer 21

Charcoal, 16 bytes

Ｆθ¿‹η0Ｆ±Ｉη←ιＦＩηι

Try it online! Link is to verbose version of code. Explanation:

Ｆθ              For each character in the input string
  ¿‹η0          If the input number is less than zero
      Ｆ±Ｉη      Repeat the negation of the input number times
          ←ι    Print the character leftwards (i.e. reversed)
      ＦＩη       Otherwise repeat the input number times
         ι      Print the character

Answer 22

Japt, 12 bytes

®pVaìr!+sVg

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Explanation

Implicit input of string U and integer V.

®pVaÃ

Map (®) each letter of U (implicitly) to itself repeated (p) abs(V) (Va) times.

¬r

Turn the string into an array of chars (¬) and reduce (r) that with...

!+sVg

"!+".slice(sign(V)) - this either reduces with + → a + b, or with !+ → b + a.
Thanks @Arnauld for the backwards-reduce idea!

Answer 23

CJam, 9 bytes

q~__*@e*%

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Answer 24

Python 3, 68 bytes

h=lambda s,n:h(s[::-1],-n)if n<0 else s[0]*n+h(s[1:],n)if s else s*n

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Answer 25

WendyScript, 46 bytes

<<f=>(s,x){<<n=""#i:s#j:0->x?x>0n+=i:n=i+n/>n}

f("Hello World", -2) // returns ddllrrooWW  oolllleeHH

Try it online!

Explanation (Ungolfed):

let f => (s, x) {
  let n = ""
  for i : s
    for j : 0->x
      if x > 0 n += i
      else n = i + n
  ret n
}

Answer 26

C89 bytes

main(int c,char**v){for(;*v[1];v[1]++)for(c=atoi(v[2]+(*v[2]=='-'));c--;)putchar(*v[1]);}

I saw Ben Perlin's version and wondered if you couldn't be shorter still and also have a full program; surely, atoi() and putchar() aren't that expensive in terms of bytes? Seems I was right!

Answer 27

Mathematica, 89 bytes

(T=Table;t=""<>T[s[[i]]~T~Abs@#2,{i,Length[s=Characters@#]}];If[#2>0,t,StringReverse@t])&

input

["Hello World!", 3]

Answer 28

QBIC, 32 bytes

g=sgn(c)[_l;||[:*g|?_sA,b*g,1|';

Explanation

            Takes inputs A$ ('Hello'), and c (-3) from the cmd line
g=sgn(c)    Save the sign of c          -1
[_l;||      FOR each char in A$
[:*g|       FOR the number of repetitions wanted    (ie: -3 * -1)
            Note that : reads a number from the cmd line, and c is the first 
            available variable to save it in after a and b got used as FOR counters.
            Also note that a negative value times the sign becomes positive.
?_s         PRINT a substring
  A         of A$
 ,b*g       startng at char n, where n is the first FOR loop counter times the sign
                That means that when c is negative, so is this. A negative starting index
                on Substring instructs QBIC to take from the right.
 ,1|        taking 1 char.
';          This bit injects a literal ; in the output QBasic, to suppress newlines om PRINT

Answer 29

Braingolf, 22 bytes

1-v{R.[v.R]v}R[v>R]v&@

Try it online!

Eeh, not bad.

Takes input as an integer and an array of characters.

Alternatively:

Braingolf, 31 bytes

l1->[M]1-v&,{R.[v.R]v}R[v>R]v&@

Try it online!

Takes input as an integer and a string

Answer 30

C, 109 bytes

char *f(int n, char *s){char *o=calloc(n,strlen(s)+1),*t=o;while(*s){for(int i=n;i--;)*t++=*s;s++;}return o;}

Starting with a function declaration that takes an int and a string and produces a string (it seems implied that memory is not preallocated and must be created) it seems that the straight-forward approach is shorter than any attempts at being cleaver that I had tried.

char *f(int n, char *s){
  char *o=calloc(n, strlen(s)+1),
    *t=o;

  while (*s) {
    for(int i=n; i--; )
      *t++=*s;
    s++;
  }

 return o;

}

Answer 31

Julia, 44 bytes

(s,x)->for i in split(s,"") print(i^abs(x))end

Answer 32

Excel VBA, _{93 89 88 86} 85 Bytes

Anonymous VBE immediate window function that takes input as string from cell [A1] and int from cell [B1] and outputs to the VBE immediate window

l=[Len(A1)]:For i=1To l:For j=1To[Abs(B1)]:?Mid([A1],IIf([B1]>0,i,l+1-i),1);:Next j,i

-4 bytes for abandoning [C1] as intermediate variable

-1 byte for adding a as intermediate variable

-2 bytes for replacing a with l, ([Len(A1)])

Answer 33

05AB1E, 9 bytes

Based off Riley's approach.

SÂΛ@²Ä×J

Try it online!

Explanation

SÂΛ@²Ä×J   Arguments: s, n
S           Push s split into individual characters
 Â          Get a (without popping) and push a reversed
  Λ        n lower than 0 (true = 1, false = 0)
    @       Get value at that index in the stack
     ²Ä×J   Repeat each character abs(n) times and join
            Implicit output

Answer 34

K/Kona, 19 18 bytes

{,/$[y>0;;|:]y#'x}

Now, in the 0 case, it does output something, but that's the empty string so I'm sure that's all gravy.

I actually missed the negative input part of this initially, which cost me eleven bytes! Definitely not my best work

Usage:

k){,/$[y>0;;|:]y#'x}["Hello World";3]
"HHHeeellllllooo   WWWooorrrlllddd!!!"
k){,/$[y>0;;|:]y#'x}["Hello World!";0]
""
k){,/$[y>0;;|:]y#'x}["String";-3]
"gggnnniiirrrtttSSS"

Answer 35

PowerShell, 89 bytes

param($s,$n)-join($s[((0..($L=$s.Length)),(-1..-$L))[$n[0]-eq45]]|%{"$_"*"$n".trim('-')})

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Generates a list of characters in the string, either forward or reversed, string-multiplies each, and joins the resulting array. $n[0]-eq45 is the ASCII code of - and .Trim('-') is shorter than [Math]::Abs($n)

Answer 36

C++, 152 bytes

#define l(x,n)for(int x=0;x<n;++x)
typedef std::string s;s f(s t,int c){s r;l(i,t.size())l(j,abs(c))r+=t[i];if(c<0)reverse(r.begin(),r.end());return r;}

And here's the full code you can test with.

#include <iostream>

#define l(x,n)for(int x=0;x<n;++x)
typedef std::string s;s f(s t,int c){s r;l(i,t.size())l(j,abs(c))r+=t[i];if(c<0)reverse(r.begin(),r.end());return r;}

int main(int argc, const char * argv[]) {
    std::cout << f("Hello World!", 3) << std::endl;
    std::cout << f("foo", 12) << std::endl;
    std::cout << f("String", -3) << std::endl;
    std::cout << f("This is a fun challenge", 0) << std::endl;
    std::cout << f("Hello\nWorld!", 2) << std::endl;
    return 0;
}

Answer 37

Brainfuck, 26 bytes

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

Try it online!

Answer 38

vim, 45 bytes

:s/\(.\)/\1<C-v>
/g
adl<esc>gJaP<esc>"add
:%norm @a
:%j!

<esc> is 0x1b, and <C-v> is 0x16.

:s/\(.\)/\1<C-v><nl>/g splits the string into one character per line.

adl<esc>gJaP<esc>"add then constructs a command in buffer a that will copy a line n times, where n is the number that was previously on this line.

%norm @a and :%j! then apply that command to each line in the file and rejoin the lines respectively.

Try it online

Answer 39

Unwanted, Unnecessary, Opportunistic, 7 bytes

IAM*|V^

Answer 40

C, 108 106 bytes

e,p,j,d;f(s,i)char*s;{p=0,e=strlen(s);for(d=i<0?p=e-1,e=-1,i=-i,-1:1;p^e;p+=d)for(j=i;j--;)putchar(s[p]);}

Answer 41

Mathematica, 56 bytes

""<>Thread@Table[Reverse[Characters@#2,2+Sign@#],Abs@#]&

Try it at the Wolfram sandbox! (Doesn't work in Mathics…)

Takes the number then the string as input.

The Reverse function can take a second parameter saying what level of the array to reverse at (so Reverse[{{1,2},{3,4}}] returns {{3,4},{1,2}} but Reverse[{{1,2},{3,4}},2] gives {{2,1},{4,3}}, for instance). When the input number is negative, we reverse at level 1, but when it's zero or positive, we reverse at level 2 or 3. Since Characters@#2 is a list with only one level, reversing at deeper levels has no effect.

Once that's done, we repeat the list of characters then transpose the resulting array using Thread so the repeated characters end up next to each other.

Answer 42

C#, 70 Bytes

using System.Linq;(s,n)=>string.Join("",s.Select(c=>new string(c,n)));

The select function is used to create a new string with the original character repeated n-times, results are joined together.

Answer 43

C#, 108 bytes

using System.Linq;s=>n=>{var a=s.Select(c=>new string(c,n>0?n:-n));return string.Concat(n<0?a.Reverse():a);}

Full/Formatted Version:

using System;
using System.Linq;

namespace TestBed
{
    class Program
    {
        static void Main(string[] args)
        {
            Func<string, Func<int, string>> f = s => n =>
            {
                var a = s.Select(c => new string(c, n > 0 ? n : -n));

                return string.Concat(n < 0 ? a.Reverse() : a);
            };

            Console.WriteLine(f("Hello World!")(3));
            Console.WriteLine(f("foo")(12));
            Console.WriteLine(f("String")(-3));
            Console.WriteLine(f("This is a fun challenge")(0));
            Console.WriteLine(f(@"Hello
World!")(2));

            Console.ReadLine();
        }
    }
}

Answer 44

tcl, 120

proc M s\ n {if $n<0 {set s [string rev $s];set n [expr -$n]};lmap c [split $s ""] {append x [string repe $c $n]};set x}

demo

Answer 45

Retina, 76 bytes

Os^$`(?<=^-.+¶.*).

^.+
$*
s`(?<=¶.*)
	
{`^1

}s`(?<=^1.*)	(.)
	$1$1
\`^¶|	

Try it online!

Answer 46

Husk, 13 bytes

So?I↔>0o`ȯṁRa

Try it online!

Ungolfed/Explanation

       o`ȯṁ    -- with the input string..
           Ra  -- ..multiply each character by the absolute value |N|
So?            -- depending on N either apply..
   I           -- ..the identity..
     >0        -- ..if N>0 or else..
    ↔          -- ..reverse it

Answer 47

Clojure, 24 bytes

#(for[c % _(range %2)]c)

Returning a string instead of a sequence of characters pushes this to 35 bytes:

#(apply str(for[c % _(range %2)]c))