<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:
        if not matrix or not matrix[0]:
            return 0

        # Initialize memoization dictionary
        seen = {}
        
        # Dimensions of the matrix
        m, n = len(matrix), len(matrix[0])

        # Find the longest increasing path starting from each cell
        longest_path = 0
        for i in range(m):
            for j in range(n):
                longest_path = max(longest_path, self.dfs(matrix, i, j, seen))
        
        return longest_path

    def inBounds(self, matrix: List[List[int]], i: int, j: int) -> bool:
        # Check if the indices are within the matrix bounds
        return 0 <= i < len(matrix) and 0 <= j < len(matrix[0])

    def dfs(self, matrix: List[List[int]], i: int, j: int, seen: dict) -> int:
        # If already computed the result for this cell, return it
        if (i, j) in seen:
            return seen[(i, j)]
        
        # Define the possible directions (up, down, left, right)
        directions = [(0, 1), (0, -1), (1, 0), (-1, 0)]
        max_path = 1  # The path length at least includes the current cell
        
        for di, dj in directions:
            ni, nj = i + di, j + dj
            if self.inBounds(matrix, ni, nj) and matrix[ni][nj] > matrix[i][j]:
                max_path = max(max_path, 1 + self.dfs(matrix, ni, nj, seen))
        
        # Memoize the result for the current cell
        seen[(i, j)] = max_path
        return max_path

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