Edit Distance

Edit Distance - Problem

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

This problem is also known as the Levenshtein Distance.

Input & Output

Example 1 — Basic Transformation

$ Input: word1 = "horse", word2 = "ros"

› Output: 3

💡 Note: horse → rorse (replace 'h' with 'r') → rose (remove 'r') → ros (remove 'e'). Total: 3 operations.

Example 2 — Insert Operations

$ Input: word1 = "intention", word2 = "execution"

› Output: 5

💡 Note: intention → inention (delete 't') → enention (replace 'i' with 'e') → exention (replace first 'n' with 'x') → exection (replace 'n' with 'c') → execution (insert 'u'). Total: 5 operations.

Example 3 — Edge Case Empty String

$ Input: word1 = "abc", word2 = ""

› Output: 3

💡 Note: Delete all 3 characters from 'abc' to get empty string. Total: 3 operations.

Constraints

0 ≤ word1.length, word2.length ≤ 500
word1 and word2 consist of lowercase English letters

Visualization

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Asked in

G Google 45 f Facebook 38 M Microsoft 32 a Amazon 28

The key insight is recognizing this as a classic dynamic programming problem where each position depends on three possible operations. The optimal approach uses a 2D DP table to build solutions bottom-up. Time: O(m×n), Space: O(m×n)

Common Approaches

✓ Dfs

⏱️ Time: N/A Space: N/A

2D Dynamic Programming

⏱️ Time: O(m×n) Space: O(m×n)

Create a 2D DP table where dp[i][j] represents minimum operations to convert word1[0..i-1] to word2[0..j-1]. Fill the table iteratively from base cases to final answer.

Recursive Brute Force

⏱️ Time: O(3^(m+n)) Space: O(m+n)

For each character position, recursively try insert, delete, and replace operations. This explores all possible transformation paths but recalculates the same subproblems multiple times.

Recursive with Memoization

⏱️ Time: O(m×n) Space: O(m×n)

Same recursive approach but stores results of subproblems in a cache (memoization table). When we encounter the same (i,j) pair again, we return the cached result instead of recalculating.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

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

int min(int a, int b, int c) {
    int result = a;
    if (b < result) result = b;
    if (c < result) result = c;
    return result;
}

int minDistance(char* word1, char* word2) {
    int m = strlen(word1);
    int n = strlen(word2);
    
    // Allocate DP table
    int** dp = (int**)malloc((m + 1) * sizeof(int*));
    for (int i = 0; i <= m; i++) {
        dp[i] = (int*)malloc((n + 1) * sizeof(int));
    }
    
    // Initialize base cases
    for (int i = 0; i <= m; i++) {
        dp[i][0] = i;
    }
    for (int j = 0; j <= n; j++) {
        dp[0][j] = j;
    }
    
    // Fill the DP table
    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] = 1 + min(dp[i-1][j],    // delete
                                  dp[i][j-1],    // insert
                                  dp[i-1][j-1]); // replace
            }
        }
    }
    
    int result = dp[m][n];
    
    // Free memory
    for (int i = 0; i <= m; i++) {
        free(dp[i]);
    }
    free(dp);
    
    return result;
}

int main() {
    char word1[1001], word2[1001];
    
    if (fgets(word1, sizeof(word1), stdin)) {
        // Remove newline
        word1[strcspn(word1, "\n")] = '\0';
    }
    
    if (fgets(word2, sizeof(word2), stdin)) {
        // Remove newline
        word2[strcspn(word2, "\n")] = '\0';
    }
    
    printf("%d\n", minDistance(word1, word2));
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

✓ Linear Growth

Space Complexity

⚡ Linearithmic Space

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