README.md

72. Edit Distance

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.

Solutions (Rust)

1. Solution

impl Solution {
    pub fn min_distance(word1: String, word2: String) -> i32 {
        let word1 = word1.as_bytes();
        let word2 = word2.as_bytes();
        let mut dp = vec![vec![i32::MAX; word2.len() + 1]; word1.len() + 1];
        dp[0] = (0..=word2.len() as i32).collect();

        for i in 1..=word1.len() {
            dp[i][0] = i as i32;
            for j in 1..=word2.len() {
                if word1[i - 1] == word2[j - 1] {
                    dp[i][j] = dp[i][j].min(dp[i - 1][j - 1]);
                }
                dp[i][j] = dp[i][j]
                    .min(dp[i][j - 1] + 1)
                    .min(dp[i - 1][j] + 1)
                    .min(dp[i - 1][j - 1] + 1);
            }
        }

        dp[word1.len()][word2.len()]
    }
}