Longest increasing path in a matrix

class Solution:
    def longestIncreasingPath(self, matrix: List[List[int]]) -> int:
        m, n = len(matrix), len(matrix[0])
        memo = [[0] * n for _ in range(m)]
        def dfs(i, j):
            if memo[i][j] != 0:
                return memo[i][j]
            for drow, dcol in [(1, 0), (0, 1), (-1, 0), (0, -1)]:
                nrow, ncol = i + drow, j + dcol
                if 0 <= nrow < m and 0 <= ncol < n:
                    if matrix[nrow][ncol] > matrix[i][j]:
                        # print((i, j), (nrow, ncol), memo[i][j])
                        memo[i][j] = max(memo[i][j], dfs(nrow, ncol))
            memo[i][j] += 1
            return memo[i][j]

        res = 0
        for i in range(m):
            for j in range(n):
            #     print(memo)
            #     print(f'start at: {(i, j)}')
                res = max(res, dfs(i, j))
                # print((i, j), res)

        return res

Longest Increasing Path in a Matrix

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