While there are many edit distance questions, such as this one, there isn't a simple question to write a program that calculates the Levenshtein distance.

Some Exposition

The Levenshtein edit distance between two strings is the minimum possible number of insertions, deletions, or substitutions to convert one word into another word. In this case, each insertion, deletion and substitution has a cost of 1.

For example, the distance between roll and rolling is 3, because deletions cost 1, and we need to delete 3 characterrs. The distance between toll and tall is 1, because substitutions cost 1.


  • The input will be two strings. You may assume that the strings are lowercase, only contain letters, are non-empty and are a maximum of 100 characters in length.
  • The output will be minimum Levenshtein edit distance of the two strings, as defined above.
  • Your code must a program or a function. It does not need to be a named function, but it cannot be a built-in function that directly computes the Levenshtein distance. Other built-ins are allowed.
  • This is code golf, so the shortest answer wins.

Some Examples

>>> lev("atoll", "bowl")
>>> lev("tar", "tarp")
>>> lev("turing", "tarpit")
>>> lev("antidisestablishmentarianism", "bulb")

As always, if the problem is unclear, please let me know. Good luck and good golfing!


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23 Answers 23


Matlab, 177 163 bytes

function l=c(a,b);m=nnz(a)+1;n=nnz(b)+1;for i=0:m-1;for j=0:n-1;z=max(i,j);try;z=min([l(i,j+1)+1,l(i+1,j)+1,l(i,j)+(a(i)~=b(j))]);end;l(i+1,j+1)=z;end;end;l=l(m,n)

This is a straightforward implementation of this formula:

enter image description here


function l=l(a,b);
for i=0:m-1;
    for j=0:n-1;
  • \$\begingroup\$ If the scored code is not what you've included, please include the scored code. Otherwise I think there's a lot of whitespace that can be golfed out. \$\endgroup\$
    – Alex A.
    Dec 23, 2015 at 17:04
  • 1
    \$\begingroup\$ @AlexA. the leading space and newlines for indentation purpose are not counted (and can be removed safely). Unce upon a time this was allowed and no one complained. \$\endgroup\$
    – edc65
    Dec 23, 2015 at 18:28
  • 1
    \$\begingroup\$ @edc65 The meta consensus now is that the code as scored should be provided. \$\endgroup\$
    – Alex A.
    Dec 23, 2015 at 18:34
  • 2
    \$\begingroup\$ Well then, the majority prefers the unreadable version. I still let the readable version here, in case anyone might want to see what is actually going on=) \$\endgroup\$
    – flawr
    Dec 23, 2015 at 20:28
  • 2
    \$\begingroup\$ It's common practice to provide both the golfed submission (the one that's scored) as well as an ungolfed version, we just require that the scored one be included. ;) \$\endgroup\$
    – Alex A.
    Dec 24, 2015 at 17:35

Pyth, 34 bytes



This is not particularly well golfed, and very slow. It can't handle anything past 2 changes in a reasonable period of time.

  • 5
    \$\begingroup\$ But it works, and that's what counts. :P \$\endgroup\$ Dec 28, 2015 at 23:54

APL (Dyalog Classic), 34 bytes

{0∊≢¨⍵:≢∊⍵⋄⌊/(≢.↑+∘∇↓¨)∘⍵¨1↓,⍳2 2}

Try it online!

  • \$\begingroup\$ This fails for 'help' f 'department' \$\endgroup\$
    – H.PWiz
    Apr 20, 2021 at 17:48
  • \$\begingroup\$ And 'look' f 'calculate' (for a different reason, I think) \$\endgroup\$
    – H.PWiz
    Apr 20, 2021 at 22:41
  • \$\begingroup\$ @H.PWiz fixed, thanks. different approach but luckily same byte count. \$\endgroup\$
    – ngn
    Apr 22, 2021 at 16:55

JavaScript (ES6) 106 113 122

Edit 16 bytes saved following @Neil suggestions

As an anonymous function.


This is a golfed implementation of the Wagner–Fischer algorithm exactly as described in the linked wikipedia article, in the section Iterative with two matrix rows (even if in fact, just 1 row is used - array w)

Less golfed

  w = [...[0,...t].keys()];
  for(i = 0; i < s.length; i++)
    w = w.map((v,j)=>
              p = j
              ? Math.min(p+1, v+1, w[j-1] + (s[i]!=t[j-1]))
              : i+1
  return p

Test snippet



[["atoll", "bowl"],["tar", "tarp"]
,["turing", "tarpit"],["antidisestablishmentarianism", "bulb"]]
.forEach(t=>console.log(t+' => '+L(...t)))
<pre id=O></pre>

  • 1
    \$\begingroup\$ Can you use [...[0,...t].keys()] instead? Saves 2 bytes if you can. \$\endgroup\$
    – Neil
    Dec 23, 2015 at 16:12
  • 1
    \$\begingroup\$ @Neil that looks ugly but it's shorter. Thx \$\endgroup\$
    – edc65
    Dec 23, 2015 at 16:41
  • 1
    \$\begingroup\$ Actually you can save another byte, [...[,...t].keys()] works too I think. \$\endgroup\$
    – Neil
    Dec 23, 2015 at 22:17
  • \$\begingroup\$ I managed to shave off another byte by using [...s].map(): (s,t)=>(w=[...[,...t].keys()],[...s].map((u,i)=>w=w.map((v,j)=>p=j--?Math.min(p,v,w[j]-(s[i-1]==t[j]))+1:i)),p) \$\endgroup\$
    – Neil
    Dec 23, 2015 at 22:28
  • \$\begingroup\$ @Neil great, thanks again! \$\endgroup\$
    – edc65
    Dec 23, 2015 at 22:44

Python 2, 151 140 138 bytes

Slow recursive implemention of the Levenshtein distance based on Wikipedia (Thanks to @Kenney for shaving of 11 chars and @Sherlock9 for another 2).

def l(s,t):
 def f(m,n):
  if m*n<1:return m or n
  return 1+min([f(m-1,n),f(m,n-1),f(m-1,n-1)-(s[m-1]==t[n-1])])
 return f(len(s),len(t))

Giving the correct answers for the presented test cases:

assert l("tar", "tarp") == 1
assert l("turing", "tarpit") == 4
assert l("antidisestablishmentarianism", "bulb") == 27        
assert l("atoll", "bowl") == 3
  • 1
    \$\begingroup\$ You might save 3-4 bytes or so by doing something like if !n*m:return n if n else m, and another 2 by return 1+min([ f(..), f(..), f(..) - (s[..] == t[..]) ]). \$\endgroup\$
    – Kenney
    Dec 24, 2015 at 21:10
  • \$\begingroup\$ You would save 2 bytes by using f(m-1,n-1)-(s[m-1]==t[n-1]) instead of f(m-1,n-1)+(s[m-1]!=t[n-1])-1. \$\endgroup\$
    – Sherlock9
    Dec 29, 2015 at 3:45
  • \$\begingroup\$ Golfed off 20 chars: codegolf.stackexchange.com/a/102910/60919 \$\endgroup\$
    – FlipTack
    Dec 11, 2016 at 14:46

Python 2, 118 bytes

A golf of this solution, but it doesn't look like Willem's been on for a year, so I'll have to post it myself:

def l(s,t):f=lambda m,n:m or n if m*n<1else-~min(f(m-1,n),f(m,n-1),f(m-1,n-1)-(s[m-1]==t[n-1]));print f(len(s),len(t))

Try on repl.it

Takes two strings and outputs the distance to STDOUT (allowed by meta). Please comment suggestions, I'm sure this can be golfed further.

  • \$\begingroup\$ Is it necessary to wrap everything in a function? Could you use two input()s or an input().split()? \$\endgroup\$
    – Sherlock9
    Dec 11, 2016 at 14:46
  • \$\begingroup\$ @Sherlock9 I tried that, but it costs 1 byte extra as far as I can tell \$\endgroup\$
    – FlipTack
    Dec 11, 2016 at 15:29
  • \$\begingroup\$ Right, I forgot that you need to define s and t somewhere in the code. Never mind. Good job :D \$\endgroup\$
    – Sherlock9
    Dec 11, 2016 at 16:27
  • 1
    \$\begingroup\$ I'm not sure why Willem used m or n. You can replace it with m+n. \$\endgroup\$
    – Arnauld
    Apr 27, 2019 at 7:05

Haskell, 67 bytes


Try it online! Example usage: "turing" # "tarpit" yields 4.

This is a recursive solution. Given two strings e and f, we first compare their first characters a and b. If they are equal, then the Levenshtein distance of e and f is the same as the Levenshtein distance of r and s, the remainder of e and f after removing the first characters. Otherwise, either a or b needs to be removed, or one is substituted by the other. [r#f,e#s,r#s] recursively computes the Levenshtein for those three cases, minimum picks the smallest one, and 1 is added to account for the necessary remove or replace operation.

If one of the strings is empty, we and up on the second line. In this case, the distance is just the length of the non-empty string, or equivalently the length of both strings concatenated together.

  • 1
    \$\begingroup\$ Wow, this is a seriously good solution, really elegant and short. \$\endgroup\$
    – ggPeti
    Mar 6, 2018 at 23:08
  • 1
    \$\begingroup\$ It seems your program outputs 0 for the input pair "abb","aab" (link). (I just found out because I wanted to steal your code for another challenge:) \$\endgroup\$
    – flawr
    Dec 29, 2019 at 21:49
  • \$\begingroup\$ @flawr Thanks for pointing out, the previous 67 byte version works correctly so I rolled back. \$\endgroup\$
    – Laikoni
    Dec 31, 2019 at 0:28

Python 3, 267 216 184 162 bytes

This function calculates the Levenshtein distance using an array that is 2 x len(word_2)+1 in size.

Edit: This doesn't get close to the Willem's Python 2 answer, but here is a more golfed answer with many little refinements here and there.

def e(p,q):
 for i in r(len(p)):
  for j in r(m):b[j+1]=1+min(a[j+1],b[j],a[j]-(p[i]==q[j]))
 return a[m]


def edit_distance(word_1,word_2):
    len_1 = len(word_1)
    len_2 = len(word_2)
    dist = [[x for x in range(len_2+1)], [1 for y in range(len_2+1)]]
    for i in range(len_1):
        for j in range(len_2):
            if word_1[i] == word_2[j]:
                dist[1][j+1] = dist[0][j]
                deletion = dist[0][j+1]+1
                insertion = dist[1][j]+1
                substitution = dist[0][j]+1
                dist[1][j+1] = min(deletion, insertion, substitution)
        dist[0], dist[1] = dist[1], [i+2 for m in range(len_2+1)]
    return dist[0][len_2]

AutoIt, 333 bytes

Func l($0,$1,$_=StringLen,$z=StringMid)
Dim $2=$_($0),$3=$_($1),$4[$2+1][$3+1]
For $5=0 To $2
For $6=0 To $3
For $5=1 To $2
For $6=1 To $3
Return $4[$2][$3]

Sample test code:

ConsoleWrite(l("atoll", "bowl") & @LF)
ConsoleWrite(l("tar", "tarp") & @LF)
ConsoleWrite(l("turing", "tarpit") & @LF)
ConsoleWrite(l("antidisestablishmentarianism", "bulb") & @LF)



k4, 66 bytes

{$[~#x;#y;~#y;#x;&/.z.s'[-1 0 -1_\:x;0 -1 -1_\:y]+1 1,~(*|x)=*|y]}

A boring and basically ungolfed impl of the algo. Ex.:

  f:{$[~#x;#y;~#y;#x;&/.z.s'[-1 0 -1_\:x;0 -1 -1_\:y]+1 1,~(*|x)=*|y]}

Seriously, 86 82 78 bytes


Hex Dump:


Try It Online

(Note that the link is to a different version because something about the online interpreter breaks with the new, shorter version, even though it works fine with the downloadable interpreter.)

It's about the most straightforward implementation Seriously allows for the recursive definition. It is very slow because it does no memoization at all. Perhaps the tabular method could be shorter (maybe by using the registers as rows), but I'm fairly happy with this, despite how much kludging-my-way-around-the-language's-shortcomings it contains. That one can use


as a proper two-argument function call was quite the nice find.


,#,#                             Read in two arguments, break them into lists of chars
    `                       `;╗ƒ put the quoted function in reg0 and immediately call it
     k;;;                        put the two lists in a list and make 3 copies
         ░                       replace the latter two with one with empty lists removed
          =                      replace two more with 1 if no empty lists removed, else 0
           "..."£@"..."£@        push the two functions described below, moving 
                                 the boolean above them both
                         I       select the correct function based on the condition
                          Ri     call the function, returning the correct distance
                                 for these substrings

   There are two functions that can be called from the main function above. Each expects 
   two strings, i and j, to be on the stack. This situation is ensured by putting 
   those strings in a list and using R to call the functions with that list as the stack.
   The first is very simple:

+l                             Concatenate the strings and take their length.
                               This is equivalent to the length of the longer
                               string, since one of the strings will be empty.

   The second function is very long and complicated. It will do the "insertion, deletion, 
   substitution" part of the recursive definition. Here's what happens in 4 parts:
│d);)                          After this, the stack is top[i-,j,i,j,ci,i-], where i- is 
                               list i with its last character, ci, chopped off.
     []oq                      this puts i- and j into a list so that they can be passed
                               as arguments recursively into the main function
         ╜Riu                  this calls the main function (from reg0) with the args
                               which will return a number to which we add 1 to get #d,
                               the min distance if we delete a character
)@d);)@                        After this, the stack is top[i,j-,ci,i-,#d,cj,j-], where 
                               j- and cj are the same idea as i- and ci
       []oq╜Riu                listify arguments, recurse and increment to get #i
                               (distance if we insert)
(@)@)@                         After this, the stack is top[i-,j-,#d,cj,#i,ci]
      []oq╜Ri                  listify arguments, recurse to get min distance between 
                               them but we still need to add 1 when we'd need to 
                               substitute because the chars we chopped off are different
(((@)                          After this, the stack is top[cj,ci,#s,#d,#i]
     =Y                        1 if they are not equal, 0 if they are
       +                       add it to the distance we find to get the distance
                               if we substitute here
        k                      put them all in a list
         m                     push the minimum distance over the three options
  • \$\begingroup\$ I like how the code tries to escape the pre-element :) \$\endgroup\$
    – user42643
    Dec 23, 2015 at 11:36

Retina, 78 72 bytes


Try it online!

In a way, this is a pure regex solution where the result is the number of positions from which the regex matches. Because why not...

Fair warning, this is super inefficient. The way this works is that it offloads the actual optimisation to the regex engine's backtracker, which simply brute forces all possible alignments, starting with as few alterations as possible and allowing one more until it's possible to match up the strings with additions, deletions and substitutions.

For a slightly more sensible solution, this one does the matching only once and doesn't have any negative lookarounds. Here, the result is the number of captures in group 2, which you could access with match.Groups[2].Captures.Count in C#, for example. It's still horribly inefficient though.


I'm explaining the second version above, because it's conceptually a bit easier (as it's just a single regex match). Here is an ungolfed version I've named the groups (or made them non-capturing) and added comments. Remember that the components in a lookbehind should be read from back to front, but alternatives and lookaheads inside those should be read from front to back. Yeah.

.+                      # Ensures backtracking from smallest to largest for next repetition
(?<ops>(?<distance>.))* # This puts the current attempted distances onto two different stacks,
                        # one to work with, and one for the result.
$                       # Make sure the lookbehind starts from the end.
(?<=                    # The basic idea is now to match up the strings character by character,
                        # allowing insertions/deletions/substitutions at the cost of one capture
                        # on <ops>. Remember to read from the bottom up.
  (?=                   # Start matching forwards again. We need to go through the other string
                        # front-to-back due to the nature of the stack (the last character we
                        # remembered from the second string must be the first character we check
                        # against in the first string).
      (?<-str>\k<str>)  # Either match the current character against the corresponding one from
                        # the other string.
      (?<-ops>          # Or consume one operation to...
        .               # consume a character without matching it to the other string (a deletion)
        (?<-str>)?      # optionally dropping a character from the other string as well 
                        # (a substitution).
    )*                  # Rinse and repeat.
    ,(?(str),)          # Ensure we reached the end of the first string while consuming all of the 
                        # second string. This is only possible if the two strings can be matched up 
                        # in no more than <distance> operations.
  ^.*,                  # Match the rest of string to get back to the front.
  (?:                   # This remembers the second string from back-to-front.
    (?<str>.)           # Either capture the current character.
    (?<-ops>.)          # Or skip it, consuming an operation. This is an insertion.

The only difference to the 72-byte version is that we can drop the leading .+ (and second group in the beginning) by finding positions at the end where we don't have enough <ops> and count all of those positions.


JavaScript (ES6),  95 91  89 bytes

Takes input as (source)(target). Essentially a port of @Willem's Python answer (later optimized by @FlipTack).


Try it online!


Python 3.8 (pre-release), 90 bytes

l=lambda a,b:min(l(A:=a[1:],B:=b[1:])-(a[0]==b[0]),l(A,b),l(a,B))+1if b>""<a else len(a+b)

Try it online!


Python 3, 105 94 93 99 bytes

-11 bytes by xnor
+6 bytes to fix equal case

l=lambda a,b:b>""<a and min(l(a[1:],b[1:])+(a[0]!=b[0]),l(a[1:],b)+1,l(a,b[1:])+1)or(a!=b)*len(a+b)

Try it online!

  • 1
    \$\begingroup\$ Nice solution. The l= needs to be included and counted because the function is recursive. You can combine the bases cases into if b>""<a else len(a+b). \$\endgroup\$
    – xnor
    Apr 27, 2019 at 22:54
  • \$\begingroup\$ Nice play with the operators, thanx! \$\endgroup\$
    – movatica
    Apr 27, 2019 at 23:38
  • 2
    \$\begingroup\$ Your solution has a problem. l("string","string") gives 2 instead of 0. This is due to min(...) or len(...) wich returns len(...) if the min is equal to 0 \$\endgroup\$
    – Jakque
    Aug 31, 2021 at 13:34
  • \$\begingroup\$ Absolutely! Fix costs 6 bytes... maybe there's room for golfing again? \$\endgroup\$
    – movatica
    Sep 8, 2021 at 20:49

Haskell, 136 bytes

Call e. Bit slow on antidisestablishmentarianism etc.

e a b=v a(l a)b(l b)
v a i b j|i*j==0=i+j|0<1=minimum[1+v a(i-1)b j,1+v a i b(j-1),fromEnum(a!!(i-1)/=b!!(j-1))+v a(i-1)b(j-1)]

GNU Prolog, 133 bytes


Takes a tuple as argument. Example of usage:

| ?- l("turing"-"tarpit",D).

D = 4


m specifies that B is A either directly or with its first character removed. d uses m as a subroutine to calculate an edit distance between the tuple elements (i.e. the distance of a series of edits that converts one into the other). Then l is a standard trick for finding the minimum of d (you take an arbitrary distance, then take an arbitrary smaller distance, repeat until you can't go smaller).


TI-Basic, 194 175 174 bytes

Prompt Str1,Str2
If YZ:Then

-19 bytes thanks to MarcMush

-1 byte by combining the 2 Prompts on the same line.

The output is stored in Ans and is displayed at the end.


Jolf, 4 bytes

Try it here!

~L   calculate the Levenshtein distance of
  i   input string
   I  and another input string

I added that builtin yesterday, but saw this challenge today, i.e., just now. Still, this answer is noncompeting.

In a newer version:


takes implicit second input.

  • \$\begingroup\$ "Your code must a program or a function. It does not need to be a named function, but it cannot be a built-in function that directly computes the Levenshtein distance. Other built-ins are allowed." \$\endgroup\$ Jun 14, 2018 at 14:17
  • \$\begingroup\$ Ah, didn't see you mentioned it's non-competing.. Better put it in the title, or add a valid program/function without builtin. \$\endgroup\$ Jun 14, 2018 at 14:19

Perl, 168 166 163 bytes

sub l{my($S,$T,$m)=(@_,100);$S*$T?do{$m=++$_<$m?$_:$m for l($S-1,$T),l($S,--$T),l(--$S,$T)-($a[$S]eq$b[$T]);$m}:$S||$T}print l~~(@a=shift=~/./g),~~(@b=shift=~/./g)

Recursive implementation. Save in a file.pl and run as perl file.pl atoll bowl.

sub l {

    ? do {
        $m = ++$_ < $m ? $_ : $m
            l($S-1,   $T),
            l($S  , --$T),
            l(--$S,   $T) - ($a[$S] eq $b[$T])
    : $S||$T
print l~~(@a=shift=~/./g),~~(@b=shift=~/./g)

The other two implementations are both longer (full matrix: 237 bytes, twoone-row iterative: 187).

  • update 166: omit () in calling l.
  • update 163: eliminate return by abusing do in trinary.

C, 192 bytes

#define m(x,y) (x>y?x:y)
#define r(x,y) l(s,ls-x,t,lt-y)
int l(char*s,int ls,char*t,int lt){if(!ls)return lt;if(!lt)return ls;a=r(1,1);if(s[ls]==t[ls])return a;return m(m(r(0,1),r(1,0)),a)+1;}


#include <stdio.h>

#define m(x,y) (x>y?x:y)
#define f(x) char x[128];fgets(x,100,stdin)
#define r(x,y) l(s,ls-x,t,lt-y)

int l(char*s,int ls,char*t,int lt)
    if(!ls) return lt;
    if(!lt) return ls;

    int a = r(1,1);
    if(s[ls]==t[ls]) return a;

    return m(m(r(0,1),r(1,0)),a)+1;

int main(void)
    printf("%d", l(a,strlen(a),b,strlen(b)));
    return 0;

C#, 215 210 198

public int L(string s,string t){int c,f,a,i,j;var v=new int[100];for(c=i=0;i<s.Length;i++)for(f=c=i,j=0;j<t.Length;j++){a=c;c=f;f=i==0?j+1:v[j];if(f<a)a=f;v[j]=c=s[i]==t[j]?c:1+(c<a?c:a);}return c;}

more readable:

public int L(string s,string t){
    int c,f,a,i,j;
    var v=new int[100];
            if (f<a) a=f;
    return c;

PowerShell v3+, 247 Bytes

$c,$d=$args;$e,$f=$c,$d|% le*;$m=[object[,]]::new($f+1,$e+1);0..$e|%{$m[0,$_]=$_};0..$f|%{$m[$_,0]=$_};1..$e|%{$i=$_;1..$f|%{$m[$_,$i]=(($m[($_-1),$i]+1),($m[$_,($i-1)]+1),($m[($_-1),($i-1)]+((1,0)[($c[($i-1)]-eq$d[($_-1)])]))|sort)[0]}};$m[$f,$e]

I ended up making this to solve another challenges involving LD's.

Code explanation with comments.

# Get both of the string passed as arguments. $c being the compare string
# and $d being the difference string. 

# Save the lengths of these strings. $e is the length of $c and $f is the length of $d
$e,$f=$c,$d|% le*

# Create the multidimentional array $m for recording LD calculations

# Populate the first column 

# Populate the first row

# Calculate the Levenshtein distance by working through each position in the matrix. 
# Working the columns
    # Save the column index for use in the next pipeline

    # Working the rows.
        # Calculate the smallest value between the following values in the matrix relative to this one
        # cell above, cell to the left, cell in the upper left. 
        # Upper left also contain the cost calculation for this pass.    
# Return the last element of the matrix to get LD 

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