<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:
        visit = set()
        self.res = 0
        ROWS, COLS = len(matrix), len(matrix[0])
        self.cache = [[1] * COLS for _ in range(ROWS)]

        neigh = [(0, -1), (-1, 0), (0, 1), (1, 0)]
        
        
        def dfs(r, c, prev):
            if (
                r < 0 or c < 0 or r > (ROWS - 1) or c > (COLS - 1) or
                matrix[r][c] <= prev
            ):
                return 0

            if (r,c) in visit:
                return self.cache[r][c]

            visit.add((r,c ))
            count=1
            re = []
            for dr, dc in neigh:
                re.append(dfs(r + dr, c + dc, matrix[r][c]))
            count += max(re)
            self.cache[r][c] = count
            self.res = max(self.res, count)
            return count
            
        
        for r in range(ROWS):
            for c in range(COLS):
                if (r, c) in visit:
                    continue
                dfs(r, c, float('-inf'))

        return self.res
25745