329-longest-increasing-path-in-a-matrix

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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

Solution(Python)

class Solution:
    def __init__(self):
        self.neighbours = [(1, 0), (-1, 0), (0, 1), (0, -1)]

    def longestIncreasingPath(self, matrix: List[List[int]]) -> int:
        return self.topdown(matrix)

    # Time Complexity: O(m*n)
    # Space Complexity: O(m*n)
    def topdown(self, matrix: List[List[int]]) -> int:
        # determine m,n
        self.m, self.n = len(matrix), len(matrix[0])
        self.max_len = 0
        self.matrix = matrix

        return max(self.dfs(i, j) for j in range(self.n) for i in range(self.m))

    @cache
    def dfs(self, i, j):

        val = self.matrix[i][j]
        return 1 + max(
            self.dfs(i - 1, j) if i and val > self.matrix[i - 1][j] else 0,
            self.dfs(i + 1, j) if i < self.m -
            1 and val > self.matrix[i + 1][j] else 0,
            self.dfs(i, j - 1) if j and val > self.matrix[i][j - 1] else 0,
            self.dfs(i, j + 1) if j < self.n -
            1 and val > self.matrix[i][j + 1] else 0,
        )