28
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Challenge

Given a string of any length which contains only digits from 0 to 9, replace each consecutive run of the digit 0 with its length.

Test Cases

  1. 12345003620004401234523623441
  2. 123450036200044123452362344
  3. 00000000000012
  4. 01234567891123456789
  5. 12345678901234567891
  6. 123456789123456789
  7. 01020300405011121324151

Note

The shortest answer in bytes wins as per rules

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2
  • 3
    \$\begingroup\$ What is a run of \$10\$ 0s replaced with? 11? \$\endgroup\$
    – Noodle9
    Commented Dec 21, 2022 at 12:28
  • 1
    \$\begingroup\$ Not OP, but I think it will be replaced with 10 as no modifications to the output were also needed, as most answers suggest \$\endgroup\$ Commented Dec 21, 2022 at 12:45

45 Answers 45

14
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Retina 0.8.2, 6 bytes

0+
$.&

Try it online! Link includes test cases. Explanation: Unary to decimal conversion.

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3
  • \$\begingroup\$ That & is neat! What does it do? \$\endgroup\$
    – Luis Mendo
    Commented Dec 21, 2022 at 12:48
  • \$\begingroup\$ @LuisMendo It refers to the entire match, like $0 does in Perl. \$\endgroup\$
    – Neil
    Commented Dec 21, 2022 at 13:04
  • \$\begingroup\$ @LuisMendo Actually my original source was wrong; PCRE, PHP and Java use $0 but Perl actually uses $&, as does JavaScript. (Retina actually supports both.) \$\endgroup\$
    – Neil
    Commented Dec 22, 2022 at 10:00
11
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JavaScript, 31 bytes

s=>s.replace(/0+/g,x=>x.length)

Try it online!

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10
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Pip, 7 bytes

-1 byte by jezza_99 / DLosc

aR+X0#_

Wow pip is really good at regex

Try It Online!

aR+X0#_
aR+X0     Replace all occurrences of (regexified zero with a +)
     #_   with it's length
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1
  • \$\begingroup\$ @DLosc might have another method, but I got this 7-byter \$\endgroup\$
    – jezza_99
    Commented Dec 21, 2022 at 22:12
9
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Japt, 7 bytes

r+iT Èl

Try it here

r+iT Èl     :Implicit input of string
r           :Replace
 +i         :  "+" prepended with
   T        :    0, giving the RegEx /0+/g
     È      :  Pass each match through a function
      l     :    Length
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8
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QuadR, 5 bytes

0+
⍵L

Try it online!

0+ Replace any runs of zeros

⍵L with the match Length

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6
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Vyxal s, 7 bytes

ĠṠƛ0cßL

Try it Online!

Explained

ĠṠƛ0cßL
Ġ       # Group on consecutive items
 Ṡ      # join each into a single string
  ƛ     # To each group:
   0c   #   if it contains 0:
     ßL #     push the length of the string
        # the s flag joins into a single string

An alternate 8 byter that uses regex match replacement

‛0+$⁽Løṙ

Try it Online!

Explained

‛0+$⁽Løṙ
‛0+       # The string "0+"
   $      # pushed under the input
    ⁽L    # a function object that returns the length of its argument
      øṙ  # replace instances of runs of 0 with their length
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6
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><>, 55 bytes

i:'0'= ?v>:0(?;o
='0':i<1<\n$v?
      \+1~  /

Try it

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6
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Python 3, 91 90 bytes

lambda x:''.join(l>"0"and l or f'{len([*g])}'for l,g in groupby(x))
from itertools import*

Try it online!

-1 byte thanks to The Thonnu

I use the itertools module's groupby function to group consecutively.

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4
  • 1
    \$\begingroup\$ You can replace the != with > to save a byte: Try it online! \$\endgroup\$
    – The Thonnu
    Commented Dec 21, 2022 at 14:07
  • 1
    \$\begingroup\$ @TheThonnu Thanks! \$\endgroup\$ Commented Dec 21, 2022 at 14:26
  • \$\begingroup\$ As this is a recursive lambda, you'll need to include the f= in your byte count. \$\endgroup\$
    – Shaggy
    Commented Dec 21, 2022 at 14:39
  • 2
    \$\begingroup\$ @Shaggy This isn't a recursive lambda, I am not calling the function within it. \$\endgroup\$ Commented Dec 21, 2022 at 14:40
6
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05AB1E, 7 bytes

0ÃηRāR:

Try it online!

  • -6 thanks to Kevin Cruijssen

Explanation

0ÃηRāR:   # Implicit input
0Ã        # List intersection with [0]
  ηR      # Reversed prefixes
    āR    # Reversed length range
      :   # Infinite replacement

Previous 13 byte answer

.γ}εD0.åig}}J   # Implicit input
.γ}             # Group by consecutive items
   ε       }    # For each group:
    D           #  Duplicate the group
     0.åi }     #  If 0 is in the group:
         g      #   Push its length
            J   # Join everything into a single string
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2
  • \$\begingroup\$ .γ} can be γ; 0.å can be _ and }} can be ] for 8 bytes: try it online. Although 0ÃηRāR: is 1 byte shorter: try it online. \$\endgroup\$ Commented Dec 28, 2022 at 8:27
  • \$\begingroup\$ @KevinCruijssen thanks \$\endgroup\$
    – The Thonnu
    Commented Dec 28, 2022 at 10:19
5
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R, 64 58 57 bytes

Edit: -1 byte thanks to @Dominic van Essen.

\(x){regmatches(x,t)=Map(attr,t<-gregexpr("0+",x),"m")
x}

Attempt This Online!

R has some weird string manipulation functions...

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2
  • \$\begingroup\$ Can't you swap the <- for simply = in the 2-liner to save a byte? \$\endgroup\$ Commented Dec 22, 2022 at 7:33
  • \$\begingroup\$ @DominicvanEssen, well, to my surprise, I can. Thanks! I thought that ...<- functions need the actual <-, but after a while of thought [<- works well with =, so why not other functions like that? \$\endgroup\$
    – pajonk
    Commented Dec 22, 2022 at 8:28
4
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Raku, 18 bytes

{S:g[0+]=$/.chars}

Try it online!

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1
  • 2
    \$\begingroup\$ still annoyed this can't be .=chars \$\endgroup\$
    – Jo King
    Commented Dec 23, 2022 at 3:55
4
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Perl 5 -p, 16 bytes

s/0+/length$&/eg

Try it online!

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4
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Vim, 28 bytes

:s/0\+/\=len(submatch(0))/g Enter

Try it online!

We cannot use 0* because it matches the empty string between digits.

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2
  • \$\begingroup\$ Here's a TIO link for you. The % isn't necessary because the input "contains only digits from 0 to 9" and therefore contains no newlines. However, if your language is Vim and not V, I think you have to count the Enter keypress after the replace command (otherwise, the replacement doesn't happen), so that makes it 28 bytes. If you want to call it V instead of Vim, you can skip the newline and claim 27 bytes. \$\endgroup\$
    – DLosc
    Commented Dec 22, 2022 at 18:22
  • \$\begingroup\$ Thanks @DLosc, edited. \$\endgroup\$ Commented Dec 22, 2022 at 23:13
3
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Retina 0.8.2, 8 bytes

(0+)
$.1

Try it online!

How it works

Match runs of zeros (0+), capture each match in a group (( )), replace it with the length of the most recent capture ($.1).

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2
  • \$\begingroup\$ Oh, I see you had already posted your answer while I was writing mine. Sorry to outgolf you like that. \$\endgroup\$
    – Neil
    Commented Dec 21, 2022 at 13:49
  • \$\begingroup\$ @Neil No problem at all, and great answer! \$\endgroup\$
    – Luis Mendo
    Commented Dec 21, 2022 at 14:13
3
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><>, 28 bytes

0i:"0"=?v{:?n}ol?!
~1+00.  >

Try it online!

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3
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Ruby, 16 bytes

gsub /0+/,&:size

Attempt This Online!

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3
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Python, 62 57 bytes

lambda n:re.sub("0+",lambda s:str(len(s[0])),n)
import re

Attempt This Online!

-5 bytes thanks to @xnor

Regex solution, port of @mathcat's Pip answer

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1
  • 2
    \$\begingroup\$ I think you can do s[0] in place of s.group() \$\endgroup\$
    – xnor
    Commented Dec 22, 2022 at 0:25
3
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Bash - 181 158 chars

Edit 1: I used this trick for the for loop, remove whitespace, condence an if statement

n=0;for ((i=0;i<${#1};i++));{ c="${1:$i:1}";if [ $c = '0' ]; then ((n++)); else [ $n -ne 0 ] && { echo -n $n;n=0;};echo -n $c;fi;};[ $n -ne 0 ] && echo -n $n

Explanation

# a counter of how far into a sequence of zeros we are
# this is zero if we're not in a 'zero-sequence'
n=0

# iterate over all the characters
# $1 is the first function argument ($0 would be the script name)
for (( i=0; i<${#1}; i++ )); do
    # get the current char
    c="${1:$i:1}"

    # if it is a zero, then increment the counter
    if [ $c = '0' ]; then
        ((n++));
    # if it is NOT a zero
    else
        # if a zero sequence is over (given that the counter is
        # not equal to zero and the current char isn't zero)
        if [ $n -ne 0 ]; then
            # print the number of zeros in the sequence,
            # and reset the counter to zero
            echo -n $n;
            n=0;
        fi;
    echo -n $c;
    fi
done
# check if there was a zero sequence terminating the string,
# as we wouldn't otherwise check as there wouldn't be a non-zero
# char initiating the check
if [ $n -ne 0 ]; then echo -n $n; fi

I referred to this SO answer for how to iterate over a string in Bash and this AU answer for a short way to increment a variable

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2
3
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Excel VBA, 100 77 bytes

Saved 23 bytes thanks to a translation by Taylor Raine

For i=2^15-1To 1Step-1:[B1]=i:[A1]=[Substitute(A1,Rept(0,B1),B1)]:Next:[B1]="

Input is in the cell A1 of the active sheet. Output is in the same cell. The code is run from the immediate window in VBA. The only clever bit is that Excel only allows 32,767 characters in a cell and counting down from there is less bytes than counting down from the length of the input.

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1
  • 1
    \$\begingroup\$ You could make this into an immediate window function something like For i=2^15-1To 1Step-1:[B1]=i:[A1]=[Substitute(A1,Rept(0,B1),B1)]:Next:[B1]=" for a pretty significant drop in bytes. (Also reminder that using ^ with no leading space makes this a 32-bit only solution as ^ is the Longlong type literal in 64-bit installs of excel) \$\endgroup\$ Commented Apr 12, 2023 at 1:12
3
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K (ngn/k), 25 bytes

{,/$(.'x)|#'x:(&~=':x)_x}

Try it online!

  • x:(&~=':x)_x split the input where it changes, and reassign to x
  • (.'x)|#'x take the maximum of the chunk of input (converted to its corresponding integer) and its count
  • ,/$ convert to strings and flatten (and implicitly return)
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3
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C (gcc), 70 bytes

n;f(char*a){*a-48?n=!printf(n?"%d":"%c",n?:*a++),*a&&f(a):f(a+!!++n);}

Try it online!

C (gcc), 62 bytes by c--

n;f(int*a){*a-48?n=!printf(n?"%d":a++,n),*a&&f(a):f(a+!!++n);}

Try it online!

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0
2
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Julia 1.0, 27 bytes

!x=replace(x,r"0+"=>length)

Try it online!

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2
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Jelly, 8 bytes

Œgċ”0ȯƊ€

A full program that accepts a string of digit characters and prints the result.

Try it online!

How?

Œgċ”0ȯƊ€ - Main Link: list of characters, S
Œg       - group runs
       € - for each group:
      Ɗ  -   last three links as a monad - f(group):
   ”0    -     literal '0' character
  ċ      -     count occurrences
     ȯ   -     logical OR (group)
         - implicit, smashing print
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2
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Husk, 8 bytes

ṁ?IosLig

Try it online!

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2
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Factor, 42 bytes

[ R/ 0+/ [ length >dec ] re-replace-with ]

Try it online!

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2
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Zsh --extendedglob, 24 bytes

<<<${1//(#m)0##/$#MATCH}

Try it Online!

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2
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Brachylog, 11 bytes

ḅ{ị0&lṫ|}ᵐc

Try it online!

Explanation

I was hoping for some automatic number -> string conversion, but no dice.

ḅ{ị0&lṫ|}ᵐc
ḅ            Break the input string into blocks of identical characters
 {      }ᵐ   Map this predicate to each block:
  ị            Convert to integer
   0           Assert that the result is zero
    &l         If so, get the length of the block
      ṫ        and convert to string
       |       If that failed (because the number wasn't zero), return the block unchanged
          c  Concatenate the results together into a single string
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2
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Java, 88 bytes

s->java.util.regex.Pattern.compile("0+").matcher(s).replaceAll(r->r.group().length()+"")

Try it online!

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2
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><>, 23 18 bytes

\&l:?n&o]
\i:'0'-?

Try it online!

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2
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Java 19, 156 161 162 bytes

interface A{static void main(String[] a){var i=0;var s="";for(var c:a[0].toCharArray())if(c==48)i++;else{s+=i>0?i+""+c:c;i=0;}System.out.print(i>0?s+i:s);}}

Without Java's golfing tax, 113 bytes

var i=0;var s="";for(var c:a[0].toCharArray())if(c==48)i++;else{s+=i>0?i+""+c:c;i=0;}System.out.print(i>0?s+i:s);

Try it online!

Edit: replaced '0' with 48 as suggested in the comments. Thanks!

Edit: String concatentation is shorter than doing var o=System.out;

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1
  • \$\begingroup\$ Welcome to Code Golf, and nice answer! \$\endgroup\$ Commented Dec 27, 2022 at 4:09

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