RUST GYM Leetcode 1092 Shortest Common Supersequence Rust Solution

1092. Shortest Common Supersequence

Given two strings str1 and str2, return the shortest string that has both str1 and str2 as subsequences. If multiple answers exist, you may return any of them.

(A string S is a subsequence of string T if deleting some number of characters from T (possibly 0, and the characters are chosen anywhere from T) results in the string S.)

Example 1:

Input: str1 = "abac", str2 = "cab"
Output: "cabac"
Explanation: 
str1 = "abac" is a subsequence of "cabac" because we can delete the first "c".
str2 = "cab" is a subsequence of "cabac" because we can delete the last "ac".
The answer provided is the shortest such string that satisfies these properties.

Note:

1 <= str1.length, str2.length <= 1000
str1 and str2 consist of lowercase English letters.

Rust Solution

struct Solution;

impl Solution {
    fn shortest_common_supersequence(str1: String, str2: String) -> String {
        let s1: Vec<char> = str1.chars().collect();
        let s2: Vec<char> = str2.chars().collect();
        let n = s1.len();
        let m = s2.len();
        let mut dp = vec![vec![(' ', 0, std::usize::MAX, std::usize::MAX); m + 1]; n + 1];
        for i in 0..n {
            dp[i + 1][0] = (s1[i], i + 1, i, 0);
        }
        for j in 0..m {
            dp[0][j + 1] = (s2[j], j + 1, 0, j);
        }
        for i in 0..n {
            for j in 0..m {
                if s1[i] == s2[j] {
                    dp[i + 1][j + 1] = (s1[i], dp[i][j].1 + 1, i, j);
                } else {
                    if dp[i][j + 1].1 < dp[i + 1][j].1 {
                        dp[i + 1][j + 1] = (s1[i], dp[i][j + 1].1 + 1, i, j + 1);
                    } else {
                        dp[i + 1][j + 1] = (s2[j], dp[i + 1][j].1 + 1, i + 1, j);
                    }
                }
            }
        }
        let mut path = vec![];
        let mut i = n;
        let mut j = m;
        while dp[i][j].0 != ' ' {
            path.push(dp[i][j].0);
            let next = dp[i][j];
            i = next.2;
            j = next.3;
        }
        path.into_iter().rev().collect()
    }
}

#[test]
fn test() {
    let str1 = "abac".to_string();
    let str2 = "cab".to_string();
    let res = "cabac".to_string();
    assert_eq!(Solution::shortest_common_supersequence(str1, str2), res);
}

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