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).

Constraints

• m == matrix.length

• n == matrix[i].length

• 1 <= m, n <= 200

• 0 <= matrix[i][j] <= 231 - 1

Approach

Links

• GeeksforGeeks

• Leetcode

• ProgramCreek

• YouTube

Examples

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

Example 3

Solutions

Solution 1

/**
 * Time complexity : 
 * Space complexity : 
 */

class Solution {
    private final int[][] dir = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}};
    
    public int longestIncreasingPath(int[][] matrix) {
        int result = 0;
        if(matrix == null || matrix.length == 0 || matrix[0].length == 0) {
            return result;
        }
        int rows = matrix.length;
        int cols = matrix[0].length;
        
        int[][] mem = new int[rows][cols];
        for(int i = 0; i < rows; i++) {
            for(int j = 0; j < cols; j++) {
                result = Math.max(result, longestIncreasingPath(matrix, mem, i, j));
            }
        }
        return result;
    }
    
    private int longestIncreasingPath(int[][] matrix, int[][] mem, int i, int j) {
        if(mem[i][j] > 0) {
            return mem[i][j];
        }
        
        for(int k = 0; k < dir.length; k++) {
            int x = i + dir[k][0];
            int y = j + dir[k][1];
            
            if(x >= 0 && x < matrix.length && y >= 0 && y < matrix[0].length && matrix[x][y] > matrix[i][j]) {
                mem[i][j] = Math.max(mem[i][j], longestIncreasingPath(matrix, mem, x, y));
            }
        }
        
        return ++mem[i][j];
    }
}

Follow up