LeetCode-in-All

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

impl Solution {
    pub fn min_distance(w1: String, w2: String) -> i32 {
        let n1 = w1.len();
        let n2 = w2.len();

        // Ensure the longer word is `w1`
        if n2 > n1 {
            return Solution::min_distance(w2, w1);
        }

        let w1_chars: Vec<char> = w1.chars().collect();
        let w2_chars: Vec<char> = w2.chars().collect();

        // Initialize dp array
        let mut dp = vec![0; n2 + 1];
        for j in 0..=n2 {
            dp[j] = j as i32;
        }

        // Dynamic programming to calculate minimum distance
        for i in 1..=n1 {
            let mut pre = dp[0];
            dp[0] = i as i32;

            for j in 1..=n2 {
                let tmp = dp[j];
                dp[j] = if w1_chars[i - 1] != w2_chars[j - 1] {
                    1 + i32::min(pre, i32::min(dp[j], dp[j - 1]))
                } else {
                    pre
                };
                pre = tmp;
            }
        }

        dp[n2]
    }
}

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