Skip to content

Edit Distance⚓︎

Link

Description⚓︎

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:
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')

Constraints:

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

Solution⚓︎

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
class Solution {
public:
    int minDistance(string word1, string word2) {
        int n = word1.length(), m = word2.length();
        word1 = ' ' + word1;
        word2 = ' ' + word2;
        vector<vector<int>> dp(n + 1, vector<int>(m + 1, 0));
        for (int i = 0; i <= n; i++) dp[i][0] = i;
        for (int i = 0; i <= m; i++) dp[0][i] = i;

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

        return dp[n][m];
    }
};