72. Edit Distance - DevCodeEx

Explanation

To solve this problem, we can use dynamic programming. We can define a 2D array dp, where dp[i][j] represents the minimum number of operations required to convert the substring word1[0...i-1] to word2[0...j-1].

We can fill in the dp array iteratively, considering the following cases:

If the characters at index i in word1 and index j in word2 are the same, then no operation is needed, and dp[i][j] = dp[i-1][j-1].
If the characters are different, we have three options:
- Insert a character: dp[i][j] = dp[i][j-1] + 1
- Delete a character: dp[i][j] = dp[i-1][j] + 1
- Replace a character: dp[i][j] = dp[i-1][j-1] + 1

The final answer will be dp[word1.length()][word2.length()].

View Leetcode Problem

class Solution {
    public int minDistance(String word1, String word2) {
        int m = word1.length();
        int n = word2.length();
        int[][] dp = new int[m + 1][n + 1];

        for (int i = 0; i <= m; i++) {
            for (int j = 0; j <= n; j++) {
                if (i == 0) {
                    dp[i][j] = j;
                } else if (j == 0) {
                    dp[i][j] = i;
                } else if (word1.charAt(i - 1) == word2.charAt(j - 1)) {
                    dp[i][j] = dp[i - 1][j - 1];
                } else {
                    dp[i][j] = 1 + Math.min(dp[i][j - 1], Math.min(dp[i - 1][j], dp[i - 1][j - 1]));
                }
            }
        }

        return dp[m][n];
    }
}

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