<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, g: List[List[int]]) -> int:
        def dfs(g,i,j,e):
            if i<0 or j<0 or i>=len(g) or j>=len(g[0]) or g[i][j]<=e:
                return 0
            if (i,j) in d:
                return d[(i,j)]
            k1=1+dfs(g,i-1,j,g[i][j])
            k2=1+dfs(g,i+1,j,g[i][j])
            k3=1+dfs(g,i,j+1,g[i][j])
            k4=1+dfs(g,i,j-1,g[i][j])
            d[(i,j)]=max(k1,k2,k3,k4)
            return max(k1,k2,k3,k4)
            
                
            
        max_len=1
        d={}
        for i in range(len(g)):
            for j in range(len(g[0])):
                
                c1=1+dfs(g,i-1,j,g[i][j])
                c2=1+dfs(g,i+1,j,g[i][j])
                c3=1+dfs(g,i,j+1,g[i][j])
                c4=1+dfs(g,i,j-1,g[i][j])
                
                if max(c1,c2,c3,c4)>max_len:
                    max_len=max(c1,c2,c3,c4)
        return max_len
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