72 Edit Distance

Given two wordsword1andword2, find the minimum number of steps required to convertword1toword2. (each operation is counted as 1 step.)

You have the following 3 operations permitted on a word:

a) Insert a character
b) Delete a character
c) Replace a character

class Solution {
    public int minDistance(String word1, String word2) {
        int[][] dp = new int[word1.length()+1][word2.length()+1];
        for(int i = 0; i <= word1.length(); i++)
            dp[i][0] = i;
        for(int i = 1; i <= word2.length(); i++)
            dp[0][i] = i;
        for (int i = 1; i < word1.length()+1; i++) {
            for (int j = 1; j < word2.length()+1; j++) {
                int insert = dp[i][j-1]+1;
                int delete = dp[i-1][j]+1;
                int replace = word1.charAt(i-1) == word2.charAt(j-1) ? dp[i-1][j-1] : dp[i-1][j-1] + 1;
                dp[i][j] = Math.min(Math.min(insert, delete), replace);
            }
        }
        return dp[word1.length()][word2.length()];
    }
}

class Solution {
    public int minDistance(String word1, String word2) {
        int[][] dp = new int[word1.length()+1][word2.length()+1];
        for(int i = 0; i <= word1.length(); i++)
            dp[i][0] = i;
        for(int i = 1; i <= word2.length(); i++)
            dp[0][i] = i;
        for (int i = 1;i <= word1.length(); i++) {
            for (int j = 1; j<= word2.length(); j++) {
                if (word1.charAt(i-1) == word2.charAt(j-1))// <--
                    dp[i][j] = dp[i-1][j-1];
                else 
                    // dp[i-1][j-1] : replace word1(i) with word2(j), because word1(0, i-1) == word2(0, j-1);
                    // dp[i  ][j-1] : delete word(j)
                    // dp[i-1][j  ] : delete word(i), because word1(0, i-1) == word2(0, j)
                    dp[i][j] = Math.min(dp[i-1][j-1], Math.min(dp[i][j-1], dp[i-1][j])) + 1; 
            }
        }
        return dp[word1.length()][word2.length()];
    }
}