<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:
        ans = 0
        visited = set()
        M = len(matrix)
        N = len(matrix[0])
        longest = {}

        # create an array which has longest inc path starting at (i, j)
        # need to pop the answer back through the dfs calls

        def dfs(i, j, prev):
            # check
            if (i < 0 or j < 0 or
                i >= M or j >= N or
                matrix[i][j] <= prev):
                return 0
            
            # if we already searched this path
            if (i, j) in longest:
                return longest[(i, j)]

            max_len = 0
            
            # search directions
            directions = [[1, 0], [-1, 0], [0, 1], [0, -1]]
            for d1, d2 in directions:
                # need to return max path
                temp = dfs(i + d1, j + d2, matrix[i][j])
                max_len = max(temp, max_len)
            
            longest[(i, j)] = 1 + max_len
            
            return 1 + max_len
        
        ans = 0

        for i in range(M):
            for j in range(N):
                temp = dfs(i, j, -1)
                ans = max(ans, temp)

        print(longest)
        return ans

        



        
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