<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:
        seen = {}
        # (prev coord, coord, len of remaing path)
        if not matrix: return 0
        m = len(matrix)
        n = len(matrix[0])
        def dfs(prev,coord,cur):

            x,y = coord
            if not (0<=x<n and 0<=y<m):
                return cur
            if prev and matrix[y][x] <= matrix[prev[1]][prev[0]]:
                return cur
            if (prev,coord) in seen:
                return cur + seen[(prev,coord)]
            path = max(
                dfs(coord,(x,y+1),cur+1),
                dfs(coord,(x,y-1),cur+1),
                dfs(coord,(x-1,y),cur+1),
                dfs(coord,(x+1,y),cur+1)
            )
            if prev:
                seen[(prev,coord)]=path-cur

            # res = max(res,path)
            return path

        res = 0
        for x in range(n):
            for y in range(m): 
          
                res = max(res,dfs(None,(x,y),0))

        return max(1,res)

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