<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>

from math import inf

class Solution:
    def longestIncreasingPath(self, matrix: List[List[int]]) -> int:
        m = len(matrix)
        n = len(matrix[0])
        dp = [[0 for _ in range(n)] for _ in range(m)]

        mi, mj = 0, 0
        memo = set()
        for i in range(m):
            for j in range(n):
                dx = [0, 0, -1, 1]
                dy = [-1, 1, 0, 0]
                ok = True
                for k in range(4):
                    ci = i + dx[k]
                    cj = j + dy[k]
                    if -1 < ci < m and -1 < cj < n:
                        if matrix[ci][cj] <= matrix[i][j]:
                            ok = False
                if ok:
                    mi, mj = i, j
                    memo.add((i, j))
                    dp[i][j] = 1

        def solve(i, j):
            if (i, j) in memo:
                return dp[i][j]
            dx = [-1, 1, 0, 0]
            dy = [0, 0, -1, 1]
            ans = 1
            for k in range(4):
                ci = i + dx[k]
                cj = j + dy[k]
                if -1 < ci < m and -1 < cj < n and matrix[ci][cj] < matrix[i][j]:
                    ans = max(ans, solve(ci, cj) + 1)
            dp[i][j] = ans
            memo.add((i, j))
            return dp[i][j]
        for i in range(m):
            for j in range(n):
                solve(i, j)
        longest = 0
        for i in range(m):
            for j in range(n):
                longest = max(longest, dp[i][j])
        return longest

                
                
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