0072. Edit Distance

Given two strings word1 and word2, return the minimum number of operations required to convert word1 to word2.

You have the following three operations permitted on a word:

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

Example 1:

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

Example 2:

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')

Constraints:

• 0 <= word1.length, word2.length <= 500
• word1 and word2 consist of lowercase English letters.

Analysis

dp[i][j]: # of ops to transfer word1[0:i] to word2[0:j]

Code

int dp_sol(string word1, string word2){
    int m = word1.size(), n = word2.size();
    vector<vector<int>> dp(m + 1, vector<int>(n + 1));
    for (int i = 0; i <= m; ++i) {
        dp[i][0] = i; // can only remove all from word2[0:i] to form an empty word1
    }

    for (int i = 0; i <= n; ++i) {
        dp[0][i] = i; // same as above
    }

    for (int i = 1; i <= m; ++i) {
        for (int j = 1; j <= n; ++j) {
            // nothing needs to be changed, because they are the same
            if (word1[i - 1] == word2[j - 1]) { 
                dp[i][j] = dp[i - 1][j - 1];
            } else { 
            // now you can either remove previous one from word1, 
            // add is equal to remove one from word2, 
            // or replace current one (no change)
                dp[i][j] = min(dp[i - 1][j - 1], min(dp[i - 1][j], dp[i][j - 1])) + 1; // +1 for the current op
            }
        }
    }
    return dp[m][n];
} 

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Last update: April 1, 2022