# 329 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`
 `````` 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``````