<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=len(matrix)
        n=len(matrix[0])
        visited=set()
        copy=[[1 for j in range(n)] for i in range(m)]
        directions=[[-1,0],[1,0],[0,-1],[0,1]]

        def dfs(i,j):
            nonlocal copy
            nonlocal visited
            res=copy[i][j]
            for d in directions:
                if 0<=i+d[0]<=m-1 and 0<=j+d[1]<=n-1 and matrix[i+d[0]][j+d[1]]<matrix[i][j]:
                    if (i+d[0],j+d[1]) not in visited:
                       dfs(i+d[0],j+d[1])
                    res=max(res,copy[i][j]+copy[i+d[0]][j+d[1]])
            copy[i][j]=res
            visited.add((i,j))
            
    
            
            
        for i in range(m):
            for j in range(n):
                x=(i,j)
                if x not in visited:
                    dfs(x[0],x[1])
                
        
        return max([max(copy[i]) for i in range(m)])
                

        
        
22755