<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:
        # Initialize the longest path counter
        longest_path = 0
        cache = {}  # To cache results of the longest path starting from each cell
        
        # Helper function for backtracking
        def recursive_backtrack(c, r, visited):
            # Check the cache to avoid recomputation
            if (c, r) in cache:
                return cache[(c, r)]
            
            # Calculate possible moves based on the current position
            possible_moves = [(c+1, r), (c-1, r), (c, r-1), (c, r+1)]
            max_path_length = 1  # Start with length 1 (the current cell itself)
            
            # Explore the possible moves (down, up, left, right)
            for move in possible_moves:
                new_c = move[0]
                new_r = move[1]
                
                # Check if the new position is within bounds and the value is increasing
                if 0 <= new_c < len(matrix[0]) and 0 <= new_r < len(matrix) and matrix[new_r][new_c] > matrix[r][c]:
                    # Recurse to the next cell and get the length of the path from there
                    path_length = 1 + recursive_backtrack(new_c, new_r, visited)
                    max_path_length = max(max_path_length, path_length)
            
            # Update the cache with the result for the current cell
            cache[(c, r)] = max_path_length
            return max_path_length
        
        # Initialize the visited array (not needed with memoization, but kept for reference)
        visited = [[0] * len(matrix[0]) for _ in range(len(matrix))]
        
        # Start backtracking from every cell and calculate the longest path
        for r in range(len(matrix)):
            for c in range(len(matrix[0])):
                longest_path = max(longest_path, recursive_backtrack(c, r, visited))
        
        return longest_path  # Return the longest path found

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