The leading zeroes challenge

Question

The challenge

Given two integers as input (x and y), output x as a string with as many leading zeroes necessary for it to be y characters long without the sign.

Rules

• If x has more than y digits, output x as string without modification.

• Output as integer is not accepted, even if there are no leading zeroes.

• When x is negative, keep the - as is and operate on the absolute value.

• Negative y should be treated as 0, meaning you output x as is (but as string)

Examples:

IN: (1,1)   OUT: "1"
IN: (1,2)   OUT: "01"
IN: (10,1)  OUT: "10"
IN: (-7,3)  OUT: "-007"
IN: (-1,1)  OUT: "-1"
IN: (10,-1) OUT: "10"
IN: (0,0)   OUT: "0"
IN: (0,1)   OUT: "0"
IN: (0,4)   OUT: "0000"

Shortest code in bytes wins, standard loopholes apply.

Answer 1

Pyth, 12 bytes

%"%0*d",+E>0

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Answer 2

Python 2, 29 bytes

lambda x,y:x.zfill(y+(x<'.'))

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Just str.zfill comes so close.

Answer 3

05AB1E, 11 10 bytes

Input given as amount_of_digits, number

ÎIÄg-×ì'-†

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Explanation

Î            # push 0 and first input
 IÄ          # push the absolute value of the second input
   g         # length
    -        # subtract, (input1-len(abs(input2))
     ×       # repeat the zero that many times
      ì      # prepend to the second input
       '-†   # move any "-" to the front

Answer 4

Python, 29 bytes

Take input as f(x,y). Using Python's % operator.

lambda x,y:'%0*d'%(y+(x<0),x)

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Answer 5

C#, 56 bytes

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a=>b=>(a<0?"-":"")+((a<0?-a:a)+"").PadLeft(b<0?0:b,'0')

Answer 6

Java (OpenJDK 8), 47 bytes

x->y->"".format("%0"+((y<1?1:y)-(x>>31))+"d",x)

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At first I thought, easy, 30 chars max (which is quite short when manipulating strings in Java). Then the exceptions happened.

Answer 7

JavaScript (ES6), 42

Recursive, parameters in reverse order, first y then x. And Currying

y=>r=x=>x<0?'-'+r(-x):`${x}`.padStart(y,0)

Test

var F=
y=>r=x=>x<0?'-'+r(-x):`${x}`.padStart(y,0)

;`IN: (1,1)   OUT: "1"
IN: (1,2)   OUT: "01"
IN: (10,1)  OUT: "10"
IN: (-7,3)  OUT: "-007"
IN: (-1,1)  OUT: "-1"
IN: (10,-1) OUT: "10"
IN: (0,0)   OUT: "0"
IN: (0,1)   OUT: "0"
IN: (0,4)   OUT: "0000"`
.split(`\n`).map(r => r.match(/[-\d]+/g))
.forEach(([x,y,k])=>{
  o = F(y)(x)
  ok = o == k
  console.log(ok?'OK':'KO',x,y,'->', o)
})

Answer 8

Python 3.6, 28 37 bytes

lambda x,y:f'{x:0{(y,0)[y<0]+(x<0)}}'

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Taking advantage of the new way to format strings in python 3.6

+9 bytes to handle y<0

Answer 9

bash, 27, 25 bytes

-2 bytes thanks to Bruce Forte

printf %0$[$2+($1<0)]d $1

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Answer 10

Husk, 12 bytes

Ö±Ωo≥⁰#±:'0s

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Explanation

Ö±Ωo≥⁰#±:'0s  Inputs are y=4 and x=-20
           s  Convert x to string: "-20"
        :'0   Prepend '0'
  Ω           until
      #±      the number of digits
   o≥⁰        is at least y: "00-20"
Ö±            Sort by is-digit: "-0020"
              Print implicitly.

Answer 11

R, 56 48 bytes

function(x,y)sprintf(paste0("%0",y+(x<0),"d"),x)

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-8 bytes thanks to djhurio

Explanation

• sprintf("%0zd",x) returns x as a string padded with zeros to be of length z
• paste0("%0",y+(x<0),"d") constructs the string "%0zd", where z is y, plus 1 if x is less than zero
• If z is less than the number of digits in x, x is printed as a string as is

Answer 12

Japt, 13 8 bytes

Takes first input (x) as a string.

®©ùTV}'-

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Saved a massive 5 bytes thanks to ETHproductions.

---

Explanation

Implicit input of string U=x and integer V=y.

® }'- splits U to an array on the minus symbol, maps over it and rejoins it to a string with a minus symbol.

© is logical AND (&&) so if the current element is truthy (a non-empty string) then pad left (ù) with 0 (T) to length V.

Answer 13

Alice, 23 bytes

/oRe./'+Ao
\I*tI&0-R$@/

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Input should be linefeed-separated with the number on the first line and the width on the second.

Explanation

/...
\...@/

This is the usual framework for linear programs in Ordinal mode. The only catch in this case is this bit:

.../...
...&...

This causes the IP to enter Cardinal mode vertically and execute just the & in Cardinal mode before resuming in Ordinal mode.

Unfolding the zigzag control flow then gives:

IRt.&'-A$o*eI/&/0+Ro@

I    Read the first line of input (the value) as a string.
R    Reverse the string.
t.   Split off the last character and duplicate it.
&    Fold the next command over this string. This doesn't really do anything,
     because the string contains only one character (so folding the next
     command is identical to executing it normally).
'-   Push "-".
A    Set intersection. Gives "-" for negative inputs and "" otherwise.
$o   If it's "-", print it, otherwise it must have been a digit which we
     leave on the stack.
*    Join the digit back onto the number. If the number was negative, this
     joins the (absolute value of the) number to an implicit empty string,
     doing nothing.
e    Push an empty string.
I    Read the width W.
/&/  Iterate the next command W times.
0    Append a zero. So we get a string of W zeros on top of the absolute
     value of the input number.
+    Superimpose. This takes the character-wise maximum of both strings
     and appends extraneous characters from the longer string. Since the
     string of zeros can never be larger than the digits in the input,
     the input itself will be uneffected, but extraneous zeros are appended,
     padding the string to the required length.
R    Reverse the result.
o    Print it.
@    Terminate the program.

Here are two alternatives, also at 23 bytes, which use Cardinal H (abs) to get rid of the -:

/R.I&0-RoH
\Ie#\'+Ao\@/

/R.H#/.+Xo
\Ie\I&0QRo@/

In principle, these are a command shorter, but the & doesn't fit into a position where there's a 1-character string on the stack, so we need to skip it with a #.

Answer 14

C, 33 bytes

f(x,y){printf("%0*d",y+(x<0),x);}

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Answer 15

Charcoal, 16 13 bytes

‹Ｎ⁰﹪⁺⁺%0ηd↔Ｉθ

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This is the shortest I could get using Charcoal without printing leading or trailing whitespaces. At least I am now starting to understand how to use the Modulo function to format strings.

The deverbosed code is as follows:

Print(Less(InputNumber(),0));    # Prints a - if the first input is less than 0
Print(Modulo(Add(Add("%0",h),"d"),Abs(Cast(q))));   # q: first input;  h: second input

• 3 bytes saved thanks to Neil!

Answer 16

Retina, 39 bytes

\d+$
$*0
((\d)*),(?<-2>-0+|0)*(0*)
$3$1

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Input should be comma-separated with the number first and the width second.

Answer 17

PHP, 45 bytes

printf("%0".($argv[2]+(0>$n=$argv[1])).d,$n);

or

[,$n,$e]=$argv;printf("%0".($e+(0>$n)).d,$n);       # requires PHP 7.1 or later

Run with -nr or try them online.

Answer 18

Mathematica, 118 bytes

(j=ToString;If[#2<=0,j@#,If[(z=IntegerLength@#)>=#2,t=z,t=#2];s=j/@PadLeft[IntegerDigits@#,t];If[#>=0,""<>s,"-"<>s]])&

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Answer 19

Mathematica, 63 62 bytes

If[#<0,"-",""]<>IntegerString[#,10,Max[#2,IntegerLength@#,1]]&

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Answer 20

Excel, 29 bytes

Using Excel's TEXT functionality ("Converts a value to text in a specific number format").

x in A1, y in B1

=TEXT(A1,REPT("0",MAX(1,B1)))

Answer 21

Octave, 44 bytes

@(x,y)fprintf(['%0',num2str(y+(x<0)),'d'],x)

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Answer 22

Haskell, 54 bytes

x#y|s<-show$abs$x=['-'|x<0]++('0'<$[length s+1..y])++s

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Answer 23

Perl 5, 22 + 1 (-n) = 23 bytes

printf"%0*d",<>+/-/,$_

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Answer 24

Ruby, 31 28 bytes

Thanks Carl for saving 3 bytes using interpolation.

->x,y{"%0#{y+(x<0?1:0)}d"%x}

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Answer 25

Japt, 14 12 bytes

Saved 2 bytes thanks to @ETHproductions

s r"%d+"_ù0V

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Answer 26

PowerShell, 25 40 bytes

param($a,$b)$a|% *g $("D$b"*($b|% *o 0))

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Explanation

This calls .ToString() on the number with a generated format string, but multiplies it by -1, 0, or 1 based on whether $b (y) is negative, 0, or positive respectively; this is to handle negative y values which format strings don't by themselves.

This seems to require wrapping negative numbers in a substatement () for it to work which is just a quirk of the invocation when using literals; if passed variables of type integer, it would not need that.

Answer 27

C# 6.0, 35 bytes

(x,y)=>(x.ToString($"D{y<0?0:y}"));

Alternative solution (51 bytes)

(x,y)=>(x.ToString(string.Format("D{0}",y<0?0:y)));

Answer 28

Clean, 90 86 83 bytes

import StdEnv
@x y=if(y<0)"-"""+++{if(c>'-')c'0'\\c<-rjustify x[k\\k<-:toString y]}

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Answer 29

C (gcc), 45 bytes

f(x,y){printf("%s%0*i","-"+(x>=0),y,abs(x));}

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Explanation

printf formats three arguments:

%s      ->    "-"+(x>=0)
%0*i    ->    y
        ->    abs(x)

%s formats the string "-"+(x>=0). "-" is really just an address, something like 41961441. In memory, this looks something like this:

MEMORY ADDRESS | 41961441  41961442 ...
VALUE          | 45 ('-')  0 (0x00) ...

When formatted into a string, C takes the address (say 41961441) and keeps on acquiring characters until a null byte (0x00) is met. When x is less than zero, the value "-"+(x>=0) has that of the original address (41961441). Otherwise, x>=0 is 1, so the expression becomes "-"+1, which points the null byte after "-", which prints nothing.

%0*i prints an integer padded with a specified number of 0s. y denotes this number. We pad abs(x) to avoid the negative in some arguments.

Answer 30

Perl, 25 + -n flag = 26 bytes

printf"%0*d",<>+($_<0),$_

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