# Levenshtein Distance

### From CodeCodex

Compute the levenshtein distance between Strings - see http://en.wikipedia.org/wiki/Levenshtein_distance

## Contents

## Implementations[edit]

### Java[edit]

public static int getLevenshteinDistance (String s, String t) { if (s == null || t == null) { throw new IllegalArgumentException("Strings must not be null"); } /* The difference between this impl. and the previous is that, rather than creating and retaining a matrix of size s.length()+1 by t.length()+1, we maintain two single-dimensional arrays of length s.length()+1. The first, d, is the 'current working' distance array that maintains the newest distance cost counts as we iterate through the characters of String s. Each time we increment the index of String t we are comparing, d is copied to p, the second int[]. Doing so allows us to retain the previous cost counts as required by the algorithm (taking the minimum of the cost count to the left, up one, and diagonally up and to the left of the current cost count being calculated). (Note that the arrays aren't really copied anymore, just switched...this is clearly much better than cloning an array or doing a System.arraycopy() each time through the outer loop.) Effectively, the difference between the two implementations is this one does not cause an out of memory condition when calculating the LD over two very large strings. */ int n = s.length(); // length of s int m = t.length(); // length of t if (n == 0) { return m; } else if (m == 0) { return n; } int p[] = new int[n+1]; //'previous' cost array, horizontally int d[] = new int[n+1]; // cost array, horizontally int _d[]; //placeholder to assist in swapping p and d // indexes into strings s and t int i; // iterates through s int j; // iterates through t char t_j; // jth character of t int cost; // cost for (i = 0; i<=n; i++) { p[i] = i; } for (j = 1; j<=m; j++) { t_j = t.charAt(j-1); d[0] = j; for (i=1; i<=n; i++) { cost = s.charAt(i-1)==t_j ? 0 : 1; // minimum of cell to the left+1, to the top+1, diagonally left and up +cost d[i] = Math.min(Math.min(d[i-1]+1, p[i]+1), p[i-1]+cost); } // copy current distance counts to 'previous row' distance counts _d = p; p = d; d = _d; } // our last action in the above loop was to switch d and p, so p now // actually has the most recent cost counts return p[n]; }

### Ruby[edit]

def levenshtein_distance(s, t) m = s.size n = t.size d = Array.new(m+1) { Array.new(n+1) } for i in 0..m d[i][0] = i end for j in 0..n d[0][j] = j end for j in 0...n for i in 0...m if s[i,1] == t[j,1] d[i+1][j+1] = d[i][j] else d[i+1][j+1] = [d[i ][j+1] + 1, # deletion d[i+1][j ] + 1, # insertion d[i ][j ] + 1 # substitution ].min end end end d[m][n] end s = "sitting" t = "kitten" p levenshtein_distance(s, t) #=> 3

### Erlang[edit]

levenshtein_distance(S, T) -> M = length(S), N = length(T), St = list_to_tuple(S), Tt = list_to_tuple(T), lists:foreach(fun(I) -> put({I, 0}, I) end, lists:seq(0, M)), lists:foreach(fun(J) -> put({0, J}, J) end, lists:seq(0, N)), lists:foreach(fun(J) -> lists:foreach(fun(I) -> levenshtein_distance(St, Tt, I, J) end, lists:seq(1, M)) end, lists:seq(1,N)), get({M, N}). levenshtein_distance(S, T, I, J) -> V = case element(I, S) =:= element(J, T) of true -> get({I-1, J-1}); false -> lists:min( [get({I-1, J }) + 1, % deletion get({I, J-1}) + 1, % insertion get({I-1, J-1}) + 1] ) % substitution end, put({I, J}, V).