72. Edit Distance

📋 Đề Bài

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.

🧠 Thuật Toán & Kỹ Thuật

Dynamic Programming (Quy hoạch động)Matrix (Ma trận)String (Chuỗi)

⏱️ Thời gian O(n×m)

💾 Không gian O(n×m)

💻 Lời Giải

C++ 0072-edit-distance.cpp

class Solution {
public:
    int minDistance(string word1, string word2) {
        int n = (int)word1.size();
        int m = (int)word2.size();

        int dp[n + 1][m + 1];

        
        for (int j = 0; j <= m; ++j) {
            dp[0][j] = j;
        }

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

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

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