Edit Distance LeetCode Solution

Problem – Edit Distance LeetCode Solution

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.

Edit Distance LeetCode Solution in Python

class Solution:
    def minDistance(self, word1, word2):
        """Dynamic programming solution"""
        m = len(word1)
        n = len(word2)
        table = [[0] * (n + 1) for _ in range(m + 1)]

        for i in range(m + 1):
            table[i][0] = i
        for j in range(n + 1):
            table[0][j] = j

        for i in range(1, m + 1):
            for j in range(1, n + 1):
                if word1[i - 1] == word2[j - 1]:
                    table[i][j] = table[i - 1][j - 1]
                else:
                    table[i][j] = 1 + min(table[i - 1][j], table[i][j - 1], table[i - 1][j - 1])
        return table[-1][-1]

Edit Distance LeetCode Solution in C++

class Solution {
public:
    int minDistance(string word1, string word2) {
        int m = word1.size(), n = word2.size();
        vector<vector<int>> dp(m + 1, vector<int>(n + 1, 0));
        for (int i = 1; i <= m; i++) {
            dp[i][0] = i;
        }
        for (int j = 1; j <= n; j++) {
            dp[0][j] = j;
        }
        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 {
                    dp[i][j] = min(dp[i - 1][j - 1], min(dp[i][j - 1], dp[i - 1][j])) + 1;
                }
            }
        }
        return dp[m][n];
    }
};

Edit Distance LeetCode Solution in Java

public class Solution {
    public int minDistance(String word1, String word2) {
        int m = word1.length();
        int n = word2.length();
        
        int[][] cost = new int[m + 1][n + 1];
        for(int i = 0; i <= m; i++)
            cost[i][0] = i;
        for(int i = 1; i <= n; i++)
            cost[0][i] = i;
        
        for(int i = 0; i < m; i++) {
            for(int j = 0; j < n; j++) {
                if(word1.charAt(i) == word2.charAt(j))
                    cost[i + 1][j + 1] = cost[i][j];
                else {
                    int a = cost[i][j];
                    int b = cost[i][j + 1];
                    int c = cost[i + 1][j];
                    cost[i + 1][j + 1] = a < b ? (a < c ? a : c) : (b < c ? b : c);
                    cost[i + 1][j + 1]++;
                }
            }
        }
        return cost[m][n];
    }
}

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