Last updated on July 20th, 2024 at 04:21 am

Here, We see Longest Increasing Path in a Matrix LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

List of all LeetCode Solution

Topics

Depth-First Search, Memorization, Topological Sort

Companies

Google

Level of Question

Hard

Longest Increasing Path in a Matrix LeetCode Solution

Problem Statement

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

1. Longest Increasing Path in a Matrix LeetCode Solution C++

class Solution {
public:
    int m, n;
    short path[200][200];
    int longestIncreasingPath(vector<vector<int>>& matrix) {
        m = matrix.size();
        n = matrix[0].size();
        memset(path, 0, sizeof(path));
        int max_path = 1;
        for(int i = 0; i < m; i++)
            for(int j = 0; j < n; j++)
                max_path = max(max_path, dfs(i, j, matrix));
        return max_path;
    }
    int dfs(int i, int j, vector<vector<int>>& mat) {
        if(path[i][j] >   0) return path[i][j];
        if(path[i][j] == -1) return 0;
        int max_next = 0;
        path[i][j] = -1;
        if(i > 0   && mat[i][j] < mat[i-1][j]) max_next = max(max_next, dfs(i-1, j, mat));
        if(j > 0   && mat[i][j] < mat[i][j-1]) max_next = max(max_next, dfs(i, j-1, mat));
        if(i < m-1 && mat[i][j] < mat[i+1][j]) max_next = max(max_next, dfs(i+1, j, mat));
        if(j < n-1 && mat[i][j] < mat[i][j+1]) max_next = max(max_next, dfs(i, j+1, mat));
        return path[i][j] = 1 + max_next;
    }

public:
    Solution() {
        ios_base::sync_with_stdio(false);
        cin.tie(nullptr);
        wcin.tie(nullptr);
        cerr.tie(nullptr);
        wcerr.tie(nullptr);
        clog.tie(nullptr);
        wclog.tie(nullptr);
    }
};

2. Longest Increasing Path in a Matrix LeetCode Solution Java

public class Solution {
    public int longestIncreasingPath(int[][] matrix) {
        if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
            return 0;
        }
        int[][] cache = new int[matrix.length][matrix[0].length];
        int max = 0;
        for (int i = 0; i < matrix.length; i++) {
            for (int j = 0; j < matrix[0].length; j++) {
                int length = findSmallAround(i, j, matrix, cache, Integer.MAX_VALUE);
                max = Math.max(length, max);
            }
        }
        return max;
    }
    private int findSmallAround(int i, int j, int[][] matrix, int[][] cache, int pre) {
        if (i < 0 || i >= matrix.length || j < 0 || j >= matrix[0].length || matrix[i][j] >= pre) {
            return 0;
        }
        if (cache[i][j] > 0) {
            return cache[i][j];
        } else {
            int cur = matrix[i][j];
            int tempMax = 0;
            tempMax = Math.max(findSmallAround(i - 1, j, matrix, cache, cur), tempMax);
            tempMax = Math.max(findSmallAround(i + 1, j, matrix, cache, cur), tempMax);
            tempMax = Math.max(findSmallAround(i, j - 1, matrix, cache, cur), tempMax);
            tempMax = Math.max(findSmallAround(i, j + 1, matrix, cache, cur), tempMax);
            cache[i][j] = ++tempMax;
            return tempMax;
        }
    }
}

3. Longest Increasing Path in a Matrix Solution JavaScript

var longestIncreasingPath = function(matrix) {
    let ylen = matrix.length, xlen = matrix[0].length, ans = 0,
        memo = Array.from({length: ylen}, el => new Uint16Array(xlen))
    const dfs = (y, x) => {
        if (memo[y][x]) return memo[y][x]
        let val = matrix[y][x]
        memo[y][x] = 1 + Math.max(
            y < ylen - 1 && matrix[y+1][x] < val ? dfs(y+1,x) : 0,
            y > 0 && matrix[y-1][x] < val ? dfs(y-1,x) : 0,
            x < xlen - 1 && matrix[y][x+1] < val ? dfs(y,x+1) : 0,
            x > 0 && matrix[y][x-1] < val ? dfs(y,x-1) : 0)
        return memo[y][x]
    }
    for (let i = 0; i < ylen; i++)
        for (let j = 0; j < xlen; j++)
            ans = Math.max(ans, dfs(i, j))
    return ans
};

4. Longest Increasing Path in a Matrix Solution Python

class Solution(object):
    def longestIncreasingPath(self, matrix):
        def dfs(i, j):
            if not dp[i][j]:
                val = matrix[i][j]
                dp[i][j] = 1 + max(
                    dfs(i - 1, j) if i and val > matrix[i - 1][j] else 0,
                    dfs(i + 1, j) if i < M - 1 and val > matrix[i + 1][j] else 0,
                    dfs(i, j - 1) if j and val > matrix[i][j - 1] else 0,
                    dfs(i, j + 1) if j < N - 1 and val > matrix[i][j + 1] else 0)
            return dp[i][j]
        if not matrix or not matrix[0]: return 0
        M, N = len(matrix), len(matrix[0])
        dp = [[0] * N for i in range(M)]
        return max(dfs(x, y) for x in range(M) for y in range(N))