Edit Distance in Java
From Wiki:
In computer science, edit distance is a way of quantifying how dissimilar two strings (e.g., words) are to one another by counting the minimum number of operations required to transform one string into the other.
There are three operations permitted on a word: replace, delete, insert. For example, the edit distance between "a" and "b" is 1, the edit distance between "abc" and "def" is 3. This post analyzes how to calculate edit distance by using dynamic programming.
Key Analysis
Let dp[i][j] stands for the edit distance between two strings with length i and j, i.e., word1[0,...,i1] and word2[0,...,j1].
There is a relation between dp[i][j] and dp[i1][j1]. Let's say we transform from one string to another. The first string has length i and it's last character is "x"; the second string has length j and its last character is "y". The following diagram shows the relation.
 if x == y, then dp[i][j] == dp[i1][j1]
 if x != y, and we insert y for word1, then dp[i][j] = dp[i][j1] + 1
 if x != y, and we delete x for word1, then dp[i][j] = dp[i1][j] + 1
 if x != y, and we replace x with y for word1, then dp[i][j] = dp[i1][j1] + 1
 When x!=y, dp[i][j] is the min of the three situations.
Initial condition:
dp[i][0] = i, dp[0][j] = j
Java Solution 1  Iteration
After the analysis above, the code is just a representation of it.
public static int minDistance(String word1, String word2) { int len1 = word1.length(); int len2 = word2.length(); // len1+1, len2+1, because finally return dp[len1][len2] int[][] dp = new int[len1 + 1][len2 + 1]; for (int i = 0; i <= len1; i++) { dp[i][0] = i; } for (int j = 0; j <= len2; j++) { dp[0][j] = j; } //iterate though, and check last char for (int i = 0; i < len1; i++) { char c1 = word1.charAt(i); for (int j = 0; j < len2; j++) { char c2 = word2.charAt(j); //if last two chars equal if (c1 == c2) { //update dp value for +1 length dp[i + 1][j + 1] = dp[i][j]; } else { int replace = dp[i][j] + 1; int insert = dp[i][j + 1] + 1; int delete = dp[i + 1][j] + 1; int min = replace > insert ? insert : replace; min = delete > min ? min : delete; dp[i + 1][j + 1] = min; } } } return dp[len1][len2]; } 
Java Solution 2  Recursion
We can write the solution in recursion.
public int minDistance(String word1, String word2) { int m=word1.length(); int n=word2.length(); int[][] mem = new int[m][n]; for(int[] arr: mem){ Arrays.fill(arr, 1); } return calDistance(word1, word2, mem, m1, n1); } private int calDistance(String word1, String word2, int[][] mem, int i, int j){ if(i<0){ return j+1; }else if(j<0){ return i+1; } if(mem[i][j]!=1){ return mem[i][j]; } if(word1.charAt(i)==word2.charAt(j)){ mem[i][j]=calDistance(word1, word2, mem, i1, j1); }else{ int prevMin = Math.min(calDistance(word1, word2, mem, i, j1), calDistance(word1, word2, mem, i1, j)); prevMin = Math.min(prevMin, calDistance(word1, word2, mem, i1, j1)); mem[i][j]=1+prevMin; } return mem[i][j]; } 
<pre><code> String foo = "bar"; </code></pre>

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