## Description

https://leetcode.com/problems/longest-increasing-path-in-a-matrix/

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 1:**

Input:nums = [ [9,9,4], [6,6,8], [2,1,1] ]Output:4Explanation:The longest increasing path is`[1, 2, 6, 9]`

.

**Example 2:**

Input:nums = [ [3,4,5], [3,2,6], [2,2,1] ]Output:4Explanation:The longest increasing path is`[3, 4, 5, 6]`

. Moving diagonally is not allowed.

## Explanation

dfs + memorization

## Python Solution

```
class Solution:
def longestIncreasingPath(self, matrix: List[List[int]]) -> int:
if not matrix:
return 0
cache = [[0 for j in range(len(matrix[0]))] for i in range(len(matrix))]
result = 0
for i in range(len(matrix)):
for j in range(len(matrix[0])):
result = max(result, self.helper(matrix, i, j, cache))
return result
def helper(self, matrix, i, j, cache):
directions = [(0, 1), (0, -1), (-1, 0), (1, 0)]
if cache[i][j] != 0:
return cache[i][j]
for direction in directions:
x = i + direction[0]
y = j + direction[1]
if 0 <= x and x < len(matrix) and 0 <= y and y < len(matrix[0]) and matrix[x][y] > matrix[i][j]:
cache[i][j] = max(cache[i][j], self.helper(matrix, x, y, cache))
cache[i][j] += 1
return cache[i][j]
```

- Time Complexity: ~MN
- Space Complexity: ~MN