<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:
          
        m, n = len(matrix), len(matrix[0])
        # Memoization table to store the longest increasing path starting from each cell
        memo = [[-1] * n for _ in range(m)]
        
        def dfs(i, j):
            # If the result for this cell is already computed, return it
            if memo[i][j] != -1:
                return memo[i][j]
            
            # Start with length 1 (the current cell itself)
            max_len = 1
            for x, y in [(-1, 0), (1, 0), (0, -1), (0, 1)]:
                ni, nj = i + x, j + y
                # Check if the new position is valid and the value is greater than the current cell
                if 0 <= ni < m and 0 <= nj < n and matrix[ni][nj] > matrix[i][j]:
                    # Explore the new cell and update the max length
                    max_len = max(max_len, 1 + dfs(ni, nj))
            
            # Memoize the result for the current cell
            memo[i][j] = max_len
            return memo[i][j]
        
        # Iterate over all cells and compute the longest increasing path starting from each one
        longest_path = 0
        for i in range(m):
            for j in range(n):
                longest_path = max(longest_path, dfs(i, j))
        
        return longest_path
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