329 Longest Increasing Path in a Matrix

🔀 Pick One

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
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24

class Solution:
    def longestIncreasingPath(self, matrix: List[List[int]]) -> int:
        m, n = len(matrix), len(matrix[0])
        cache = collections.defaultdict(int) # [int, int] => int

        def dfs(i, j):
            if (i, j) in cache:
                return cache[(i, j)]
            for dx, dy in [(1, 0), (-1, 0), (0, 1), (0, -1)]:
                x, y = i + dx, j + dy
                if (
                    0 <= x < m and
                    0 <= y < n and
                    matrix[x][y] > matrix[i][j]
                ):
                    cache[(i, j)] = max(cache[(i, j)], dfs(x, y))
            cache[(i, j)] += 1
            return cache[(i, j)]
        
        ans = 0
        for i in range(m):
            for j in range(n):
                ans = max(ans, dfs(i, j))
        return ans