<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, grid: List[List[int]]) -> int:
        m,n=len(grid),len(grid[0])
        # directions = [0, 1, 0, -1, 0] # four directions 
        directions = [(0,1),(0,-1),(1,0),(-1,0)]
        
        @lru_cache(maxsize=None) # using python cache lib for memoization
        def dfs(r,c):
            ans=1                  
			# iterating through all 4 directions
            for x,y in directions: 
                new_row,new_col=r+x, c+y # calculating the new cell
				# check if new cell is within the grid bounds and is an increasing sequence
                if 0<=new_row<m and 0<=new_col<n and grid[new_row][new_col]>grid[r][c]: 
                    ans = max(ans, dfs(new_row, new_col) + 1 )  # finding the max length of valid path from the current cell                                      
            return ans
        
        return max(dfs(r,c) for r in range(m) for c in range(n)) 
27175