<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:
        h = len(matrix)
        w = len(matrix[0])

        dp = [[-1 for _ in range(w)] for _ in range(h)]  # Initialize with -1 to indicate unvisited
        dc = [[0, 1], [0, -1], [1, 0], [-1, 0]]

        def dfs(i, j, pr = -1):
            if i < 0 or i >= h or j < 0 or j >= w or matrix[i][j] <= pr:
                return 0  # Invalid move, so return 0 as no path

            if dp[i][j] != -1:
                return dp[i][j]  # Return the cached value if it exists

            max_path = 1  # Start with length 1 (the current cell)
            
            for d in dc:
                max_path = max(max_path, 1 + dfs(i + d[0], j + d[1], matrix[i][j]))

            dp[i][j] = max_path  # Cache the result in dp table
            return max_path
        
        res = 0
        for i in range(h):
            for j in range(w):
                res = max(res, dfs(i, j))
        
        return res

19115