RUST GYM Leetcode 329 Longest Increasing Path in a Matrix Rust Solution

329. Longest Increasing Path in a Matrix

Given an m x n matrix, return the length of the longest increasing path in matrix.

From each cell, you can either move in four directions: left, right, up, or down. You may not move diagonally or move outside the boundary (i.e., wrap-around is not allowed).

Example 1:

Input: matrix = [[9,9,4],[6,6,8],[2,1,1]]
Output: 4
Explanation: The longest increasing path is [1, 2, 6, 9].

Example 2:

Input: matrix = [[3,4,5],[3,2,6],[2,2,1]]
Output: 4
Explanation: The longest increasing path is [3, 4, 5, 6]. Moving diagonally is not allowed.

Example 3:

Input: matrix = [[1]]
Output: 1

Constraints:

m == matrix.length
n == matrix[i].length
1 <= m, n <= 200
0 <= matrix[i][j] <= 2³¹ - 1

Rust Solution

struct Solution;

impl Solution {
    fn longest_increasing_path(matrix: Vec<Vec<i32>>) -> i32 {
        let n = matrix.len();
        if n == 0 {
            return 0;
        }
        let m = matrix[0].len();
        let mut memo = vec![vec![0; m]; n];
        let mut res = 0;
        for i in 0..n {
            for j in 0..m {
                if memo[i][j] == 0 {
                    memo[i][j] = Self::dfs(i, j, &mut memo, &matrix);
                    res = res.max(memo[i][j]);
                }
            }
        }
        res
    }

    fn dfs(i: usize, j: usize, memo: &mut Vec<Vec<i32>>, matrix: &[Vec<i32>]) -> i32 {
        if memo[i][j] != 0 {
            return memo[i][j];
        }
        let n = matrix.len();
        let m = matrix[0].len();
        let mut res = 1;
        if i > 0 && matrix[i - 1][j] > matrix[i][j] {
            res = res.max(Self::dfs(i - 1, j, memo, matrix) + 1);
        }
        if j > 0 && matrix[i][j - 1] > matrix[i][j] {
            res = res.max(Self::dfs(i, j - 1, memo, matrix) + 1);
        }
        if i + 1 < n && matrix[i + 1][j] > matrix[i][j] {
            res = res.max(Self::dfs(i + 1, j, memo, matrix) + 1);
        }
        if j + 1 < m && matrix[i][j + 1] > matrix[i][j] {
            res = res.max(Self::dfs(i, j + 1, memo, matrix) + 1);
        }
        memo[i][j] = res;
        res
    }
}

#[test]
fn test() {
    let matrix = vec_vec_i32![[9, 9, 4], [6, 6, 8], [2, 1, 1]];
    let res = 4;
    assert_eq!(Solution::longest_increasing_path(matrix), res);
}

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