LeetCode-72-Edit-Distance.java

/*
https://discuss.leetcode.com/topic/17639/20ms-detailed-explained-c-solutions-o-n-space
https://www.cnblogs.com/grandyang/p/4344107.html
https://www.youtube.com/watch?v=Q4i_rqON2-E

Example: 

a b c d e f
a z c e d

b -> z
delete d
f -> d

Total 3 edits to convert "abcdef" to "azced"

Three Operations:

Case 1: word1[i] = word2[j]
f(i, j) = f(i - 1, j - 1)

Case 2: word1[i] != word2[j], then we must either insert, delete or replace, whichever is cheaper

1. f(i, j) = f(i, j - 1) + 1    -   insert in word2
2. f(i, j) = f(i - 1, j) + 1    -   delete in word1
3. f(i, j) = f(i - 1, j - 1) + 1    -   replace the char



Matrix
    a b c d e f 
  0 1 1 3 4 5 6
a 1 0 1 2 3 4 5
z 2 1 1 2 3 4 5
c 3 2 2 1 2 3 4
e 4 3 3 2 2 2 4
d 5 4 4 3 3 3 3
*/
public class Solution {
//     public int minDistance(String word1, String word2) {
//         int m = word1.length(), n = word2.length();
        
//         // state matrix
//         int[][] dp = new int[m + 1][n + 1];
        
//         // init matrix
//         for(int i = 1; i <= m; i++) dp[i][0] = i;
//         for(int j = 1; j <= n; j++) dp[0][j] = j;
        
//         // calculate matrix
//         for(int i = 0; i < m; i++){
//             for(int j = 0; j < n; j++){
//                 if(word1.charAt(i) == word2.charAt(j)) dp[i + 1][j + 1] = dp[i][j];
//                 else dp[i + 1][j + 1] = Math.min(dp[i][j] + 1, Math.min(dp[i][j + 1] + 1, dp[i + 1][j] + 1));
//             }
//         }
        
//         return dp[m][n];
//     }
    
    // similar to the above one
    public int minDistance(String word1, String word2) {
        int m = word1.length(), n = word2.length();
        
        // subproblem
        int[][] dp = new int[m + 1][n + 1];
        
        // init
        for (int i = 1; i <= m; i++) dp[i][0] = i;
        for (int j = 1; j <= n; j++) dp[0][j] = j;
        
        // recurrence relation
        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] = Math.min(dp[i - 1][j - 1], Math.min(dp[i - 1][j], dp[i][j - 1])) + 1;
                }
            }
        }
        
        // ans
        return dp[m][n];
    }
}