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Double Sequence III

Description

Given two words word1 and word2, find the minimum number of operations required to convert word1 to word2.

You have the following 3 operations permitted on a word:

• Insert a character
• Delete a character
• Replace a character

Question Link

https://leetcode.com/problems/edit-distance/description/

Example I:

Input: word1 = "horse", word2 = "ros"
Output: 3
Explanation:
horse -> rorse (replace 'h' with 'r')
rorse -> rose (remove 'r')
rose -> ros (remove 'e')

Example II:

Input: word1 = "intention", word2 = "execution" Output: 5
Explanation:
intention -> inention (remove 't')
inention -> enention (replace 'i' with 'e') enention -> exention (replace 'n' with 'x') exention -> exection (replace 'n' with 'c') exection -> execution (insert 'u')

Questions

• LeetCode - 72. Edit Distance

Transform Function

for(int i = 1; i <= m; i++) {
    for(int j = 1; j <= n; j++) {
        if(word1.charAt(i - 1) == word2.charAt(j - 1)) dp[i][j] = dp[i - 1][j - 1];
        else dp[i][j] = 1 + Math.min(dp[i - 1][j - 1], Math.min(dp[i - 1][j], dp[i][j - 1]));
    }
}

Solution

class Solution {
    public int minDistance(String word1, String word2) {
        char[] c1 = word1.toCharArray(), c2 = word2.toCharArray();
        int m = c1.length, n = c2.length;
        int[][] dp = new int[m + 1][n + 1];
        for(int i = 0; i <= m; i++) dp[i][0] = i;
        for(int j = 0; j <= n; j++) dp[0][j] = j;

        for(int i = 1; i <= m; i++) {
            for(int j = 1; j <= n; j++) {
                if(c1[i - 1] == c2[j - 1]) dp[i][j] = dp[i - 1][j - 1];
                else {
                    dp[i][j] = 1 + Math.min(dp[i - 1][j - 1], Math.min(dp[i - 1][j], dp[i][j - 1]));
                }
            }
        }
        return dp[m][n];
    }
}