329. Longest Increasing Path in a Matrix

Description

Given an m x n integers 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

Solutions

Solution 1: Memoization Search

We design a function $dfs(i, j)$, which represents the length of the longest increasing path that can be obtained starting from the coordinate $(i, j)$ in the matrix. The answer is $\max_{i, j} \textit{dfs}(i, j)$.

The execution logic of the function $dfs(i, j)$ is as follows:

If $(i, j)$ has been visited, directly return $\textit{f}(i, j)$;
Otherwise, search $(i, j)$, search the coordinates $(x, y)$ in four directions. If $0 \le x < m, 0 \le y < n$ and $matrix[x][y] > matrix[i][j]$, then search $(x, y)$. After the search is over, update $\textit{f}(i, j)$ to $\textit{f}(i, j) = \max(\textit{f}(i, j), \textit{f}(x, y) + 1)$. Finally, return $\textit{f}(i, j)$.

The time complexity is $O(m \times n)$, and the space complexity is $O(m \times n)$. Where $m$ and $n$ are the number of rows and columns of the matrix, respectively.

Similar problems:

2328. Number of Increasing Paths in a Grid

Python3JavaC++GoTypeScript

class Solution:
    def longestIncreasingPath(self, matrix: List[List[int]]) -> int:
        @cache
        def dfs(i: int, j: int) -> int:
            ans = 0
            for a, b in pairwise((-1, 0, 1, 0, -1)):
                x, y = i + a, j + b
                if 0 <= x < m and 0 <= y < n and matrix[x][y] > matrix[i][j]:
                    ans = max(ans, dfs(x, y))
            return ans + 1

        m, n = len(matrix), len(matrix[0])
        return max(dfs(i, j) for i in range(m) for j in range(n))

class Solution {
    private int m;
    private int n;
    private int[][] matrix;
    private int[][] f;

    public int longestIncreasingPath(int[][] matrix) {
        m = matrix.length;
        n = matrix[0].length;
        f = new int[m][n];
        this.matrix = matrix;
        int ans = 0;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                ans = Math.max(ans, dfs(i, j));
            }
        }
        return ans;
    }

    private int dfs(int i, int j) {
        if (f[i][j] != 0) {
            return f[i][j];
        }
        int[] dirs = {-1, 0, 1, 0, -1};
        for (int k = 0; k < 4; ++k) {
            int x = i + dirs[k];
            int y = j + dirs[k + 1];
            if (x >= 0 && x < m && y >= 0 && y < n && matrix[x][y] > matrix[i][j]) {
                f[i][j] = Math.max(f[i][j], dfs(x, y));
            }
        }
        return ++f[i][j];
    }
}

class Solution {
public:
    int longestIncreasingPath(vector<vector<int>>& matrix) {
        int m = matrix.size(), n = matrix[0].size();
        int f[m][n];
        memset(f, 0, sizeof(f));
        int ans = 0;
        int dirs[5] = {-1, 0, 1, 0, -1};

        function<int(int, int)> dfs = [&](int i, int j) -> int {
            if (f[i][j]) {
                return f[i][j];
            }
            for (int k = 0; k < 4; ++k) {
                int x = i + dirs[k], y = j + dirs[k + 1];
                if (x >= 0 && x < m && y >= 0 && y < n && matrix[x][y] > matrix[i][j]) {
                    f[i][j] = max(f[i][j], dfs(x, y));
                }
            }
            return ++f[i][j];
        };

        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                ans = max(ans, dfs(i, j));
            }
        }
        return ans;
    }
};

func longestIncreasingPath(matrix [][]int) (ans int) {
    m, n := len(matrix), len(matrix[0])
    f := make([][]int, m)
    for i := range f {
        f[i] = make([]int, n)
    }
    dirs := [5]int{-1, 0, 1, 0, -1}
    var dfs func(i, j int) int
    dfs = func(i, j int) int {
        if f[i][j] != 0 {
            return f[i][j]
        }
        for k := 0; k < 4; k++ {
            x, y := i+dirs[k], j+dirs[k+1]
            if 0 <= x && x < m && 0 <= y && y < n && matrix[x][y] > matrix[i][j] {
                f[i][j] = max(f[i][j], dfs(x, y))
            }
        }
        f[i][j]++
        return f[i][j]
    }
    for i := 0; i < m; i++ {
        for j := 0; j < n; j++ {
            ans = max(ans, dfs(i, j))
        }
    }
    return
}

function longestIncreasingPath(matrix: number[][]): number {
    const m = matrix.length;
    const n = matrix[0].length;
    const f: number[][] = Array(m)
        .fill(0)
        .map(() => Array(n).fill(0));
    const dirs = [-1, 0, 1, 0, -1];
    const dfs = (i: number, j: number): number => {
        if (f[i][j] > 0) {
            return f[i][j];
        }
        for (let k = 0; k < 4; ++k) {
            const x = i + dirs[k];
            const y = j + dirs[k + 1];
            if (x >= 0 && x < m && y >= 0 && y < n && matrix[x][y] > matrix[i][j]) {
                f[i][j] = Math.max(f[i][j], dfs(x, y));
            }
        }
        return ++f[i][j];
    };
    let ans = 0;
    for (let i = 0; i < m; ++i) {
        for (let j = 0; j < n; ++j) {
            ans = Math.max(ans, dfs(i, j));
        }
    }
    return ans;
}

329. Longest Increasing Path in a Matrix

Description

Solutions

Solution 1: Memoization Search

Comments