Replace 0s In a String With Their Consecutive Counts

Question

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. 1234500362000440 → 1234523623441
2. 123450036200044 → 123452362344
3. 000000000000 → 12
4. 0123456789 → 1123456789
5. 1234567890 → 1234567891
6. 123456789 → 123456789
7. 010203004050 → 11121324151

Note

The shortest answer in bytes wins as per code-golf rules

Answer 1

Retina 0.8.2, 6 bytes

0+
$.&

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

Answer 2

JavaScript, 31 bytes

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

Try it online!

Answer 3

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

Answer 4

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

Answer 5

QuadR, 5 bytes

0+
⍵L

Try it online!

0+ Replace any runs of zeros

⍵L with the match Length

Answer 6

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

Answer 7

><>, 55 bytes

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

Try it

Answer 8

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.

Answer 9

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

Answer 10

Perl 5 -p, 16 bytes

s/0+/length$&/eg

Try it online!

Answer 11

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...

Answer 12

Raku, 18 bytes

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

Try it online!

Answer 13

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.

Answer 14

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).

Answer 15

><>, 28 bytes

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

Try it online!

Answer 16

Ruby, 16 bytes

gsub /0+/,&:size

Attempt This Online!

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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.

Answer 21

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)

Answer 22

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!

Answer 23

Julia 1.0, 27 bytes

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

Try it online!

Answer 24

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

Answer 25

Husk, 8 bytes

ṁ?IosLig

Try it online!

Answer 26

Factor, 42 bytes

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

Try it online!

Answer 27

Zsh --extendedglob, 24 bytes

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

Try it Online!

Answer 28

Java, 88 bytes

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

Try it online!

Answer 29

><>, 23 18 bytes

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

Try it online!

Answer 30

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;