<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>

import collections
class Solution:

    def dfs(self,r,c):
        m=self.m
        n=self.n
        matrix=self.matrix
        val=matrix[r][c]
        if r >= m or r<0 or c >=n or c<0:
            return 0

        if self.nodes[r][c]>0:
            return self.nodes[r][c]

        moves=[[r+1,c],[r-1,c],[r,c+1],[r,c-1]]

        curr_max=0

        for i in range(4): # 4 moves
            next_r,next_c = moves[i]
            if next_r>=0 and next_r<m and next_c>=0 and next_c<n and matrix[next_r][next_c]>val:
                curr_max=max(curr_max, self.dfs(next_r,next_c))
            

        self.nodes[r][c]=curr_max+1

        if curr_max+1>self.max_path:
            self.max_path=curr_max+1

        return curr_max+1
         
        
    
    def longestIncreasingPath(self, matrix: List[List[int]]) -> int:
        # solution idea: dfs with each matrix block as a node
        m=self.m=len(matrix) # rows 
        n=self.n=len(matrix[0]) # columns 
        self.matrix=matrix
        self.nodes=[[0 for i in range(self.n)] for i in range(self.m)]
        self.max_path=0
        for i in range(m*n):
            self.dfs(i//n, i%n) # pass in row and column 

        return self.max_path
            
                
                
                


                



        
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