# LeetCode 329. Longest Increasing Path in a Matrix

## 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: 4
Explanation: The longest increasing path is `[1, 2, 6, 9]`.
```

Example 2:

```Input: nums =
[
[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.```

## 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))] for i in range(len(matrix))]

result = 0

for i in range(len(matrix)):
for j in range(len(matrix)):
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
y = j + direction

if 0 <= x and x < len(matrix) and 0 <= y and y < len(matrix) 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