<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 class="example">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 class="example">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 class="example">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>

class Solution:
    def longestIncreasingPath(self, matrix: List[List[int]]) -> int:
        max_path = 0
        visited = defaultdict(int)

        def dfs(i, j, prev, curr_visited):
            if i < 0 or j < 0 or i >= len(matrix) or j >= len(matrix[0]):
                return 0
            if matrix[i][j] <= prev:
                return 0
            if (i, j) in curr_visited:
                return 0
            if (i, j) in visited:
                return visited[(i, j)]
            curr_visited.add((i,j))
            directions = [(0,1),(0,-1),(1,0),(-1,0)]
            max_val = 0
            for x, y in directions:
                max_val = max(max_val, dfs(i+x, j+y, matrix[i][j], curr_visited))
            curr_visited.remove((i,j))
            visited[(i,j)] = 1 + max_val
            return visited[(i,j)]

        for i in range(len(matrix)):
            for j in range(len(matrix[0])):
                if (i, j) not in visited:
                    curr_visited = set()
                    length = dfs(i, j, -1, curr_visited)
                    max_path = max(max_path, length)
        return max_path  
24185