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


Try it on <a href='https://leetcode.com/problems/329-longest-increasing-path-in-a-matrix'>leetcode</a>

## Description
<div class="description">
<div><p>Given an <code>m x n</code> integers <code>matrix</code>, return <em>the length of the longest increasing path in </em><code>matrix</code>.</p>

<p>From each cell, you can either move in four directions: left, right, up, or down. You <strong>may not</strong> move <strong>diagonally</strong> or move <strong>outside the boundary</strong> (i.e., wrap-around is not allowed).</p>

<p>&nbsp;</p>
<p><strong>Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/01/05/grid1.jpg" style="width: 242px; height: 242px;">
<pre><strong>Input:</strong> matrix = [[9,9,4],[6,6,8],[2,1,1]]
<strong>Output:</strong> 4
<strong>Explanation:</strong> The longest increasing path is <code>[1, 2, 6, 9]</code>.
</pre>

<p><strong>Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/01/27/tmp-grid.jpg" style="width: 253px; height: 253px;">
<pre><strong>Input:</strong> matrix = [[3,4,5],[3,2,6],[2,2,1]]
<strong>Output:</strong> 4
<strong>Explanation: </strong>The longest increasing path is <code>[3, 4, 5, 6]</code>. Moving diagonally is not allowed.
</pre>

<p><strong>Example 3:</strong></p>

<pre><strong>Input:</strong> matrix = [[1]]
<strong>Output:</strong> 1
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == matrix.length</code></li>
	<li><code>n == matrix[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 200</code></li>
	<li><code>0 &lt;= matrix[i][j] &lt;= 2<sup>31</sup> - 1</code></li>
</ul>
</div>
</div>

## Solution(Python)
```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,
        )

```