Problem 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
Example 1:

Input: word1 = “horse”, word2 = “ros”
Output: 3
Explanation:

1
2
3
horse -> rorse (replace 'h' with 'r')
rorse -> rose (remove 'r')
rose -> ros (remove 'e')

Example 2:

Input: word1 = “intention”, word2 = “execution”
Output: 5
Explanation:

1
2
3
4
5
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')

Solution:

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
class Solution { 
public:
    int minDistance(string word1, string word2) { 
        int m = word1.length(), n= word2.length();
        vector<vector<int>> dp(m+1, vector<int>(n+1, 0));
        
        //convert word2 to word1 if word2 is empty, so the operation is m times
        //ex: word1: "ABCDE", word2= ""  
        //notice: start from 1 because dp[i][j] denotes word1[0...i-1] are the same as word2[0...j-1]
        for(int i= 1; i<= m; ++i)
            dp[i][0] = i;
        
        //convert word1 to word2 if word1 is empty, so the operation is n times
        for(int i= 1; i<= n; ++i)
            dp[0][i] = i;
        
        
        for(int i = 1; i<= m; ++i){
            for(int j= 1; j<= n; ++j){
                if(word1[i-1] == word2[j-1])
                    dp[i][j] = dp[i-1][j-1];
                else
                    //replace: means till the dp[i-1][j-1] is the same, add 1 operation by replacement
                    //  example: word1: "ABCDE", word2: "ABCDF" 
                    //           therefore, only needs to 1 move based on dp[i-1][j-1]
                    
                    //deletion: the question is convert from word1 to word2, 
                    //          means till the dp[i-1][j] is the same, add 1 ...
                    //  example: word1: "ABCDE", word2: "ABCD". i=4 j=3
                    //           add 1 move based on dp[i-1][j]
                    
                    //insertion: the question is convert from word1 to word2, 
                    //          means till the dp[i][j-1] is the same, add 1 ...
                    //  example: word1: "ABCD", word2: "ABCDE". i=3 j=4
                    //           add 1 move based on dp[i][j-1]
                    dp[i][j] = min(dp[i-1][j-1]+1, min(dp[i-1][j]+1, dp[i][j-1]+1));
            }
        }
        
        return dp[m][n];
    }
};