LeetCode in C

72. Edit Distance

Hard

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.

Solution

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int minDistance(char* w1, char* w2) {
    int n1 = strlen(w1);
    int n2 = strlen(w2);
    
    // Ensure w1 is the longer string
    if (n2 > n1) {
        return minDistance(w2, w1);
    }
    
    // Create a DP array to store the distances
    int* dp = (int*)malloc((n2 + 1) * sizeof(int));
    
    // Initialize the DP array
    for (int j = 0; j <= n2; j++) {
        dp[j] = j;
    }

    // Fill the DP table
    for (int i = 1; i <= n1; i++) {
        int pre = dp[0];
        dp[0] = i; // First column represents distance to w1
        for (int j = 1; j <= n2; j++) {
            int tmp = dp[j];
            dp[j] = (w1[i - 1] != w2[j - 1]) 
                ? 1 + (pre < dp[j] ? (pre < dp[j - 1] ? pre : dp[j - 1]) : (dp[j] < dp[j - 1] ? dp[j] : dp[j - 1]))
                : pre;
            pre = tmp; // Store the previous value
        }
    }

    int result = dp[n2];

    // Free allocated memory
    free(dp);

    return result;
}

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