[LeetCode] Problem 329 - Longest Increasing Path in a Matrix

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Given an integer matrix, find the length of the longest increasing path.

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

Example

No.1

Input: nums = 

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2
3
4
5
[
  [9,9,4],
  [6,6,8],
  [2,1,1]
]

Output: 4

Explanation: The longest increasing path is [1, 2, 6, 9].

No.2

Input: nums = 

1
2
3
4
5
[
  [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.

Code

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private int[][] dirs = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};

public int longestIncreasingPath(int[][] matrix) {
    int result = 0;
    
    if (matrix == null || matrix.length == 0 || matrix[0].length == 0)
        return result;
    
    int m = matrix.length;
    int n = matrix[0].length;
    int[][] path = new int[m][n];

    for (int i = 0; i < m; i++) {
        for (int j = 0; j < n; j++)
            result = Math.max(result, dfs(matrix, path, m, n, i, j));
    }

    return result;
}

private int dfs(int[][] matrix, int[][] path, int m, int n, int x, int y) {
    if (path[x][y] != 0)
        return path[x][y];

    path[x][y] = 1;
    
    for (int[] dir : dirs) {
        int newX = x + dir[0];
        int newY = y + dir[1];

        if (newX < 0 || newY < 0 || newX >= m || newY >= n)
            continue;
        
        if (matrix[x][y] >= matrix[newX][newY])
            continue;
        
        path[x][y] = Math.max(path[x][y], 1 + dfs(matrix, path, m, n, newX, newY));
    }

    return path[x][y];
}