<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:
        """
        a cell in the matrix as a node,
a directed edge from node x to node y if x and y are adjacent and x's value < y's value
Then a graph is formed.

No cycles can exist in the graph, i.e. a DAG is formed.

The problem becomes to get the longest path in the DAG.

Topological sort can iterate the vertices of a DAG in the linear ordering.

Using Kahn's algorithm(BFS) to implement topological sort while counting the levels can give us the longest chain of nodes in the DAG.
        """
        m, n = len(matrix), len(matrix[0])
        ind = [[0] * n for _ in range(m)]
        dirs = [(1, 0), (0, 1), (-1, 0), (0, -1)]
        for i in range(m):
            for j in range(n):
                for dx, dy in dirs:
                    ni, nj = i + dx, j + dy
                    if 0 <= ni < m and 0 <= nj < n and matrix[ni][nj] > matrix[i][j]:
                        ind[ni][nj] += 1
        
        q = deque([(x, y) for x in range(m) for y in range(n) if ind[x][y] == 0])
        max_path = 0
        while q:
            max_path += 1
            for _ in range(len(q)):
                i, j = q.popleft()
                for dx, dy in dirs:
                    ni, nj = i + dx, j + dy
                    if 0 <= ni < m and 0 <= nj < n and matrix[ni][nj] > matrix[i][j]:
                        ind[ni][nj] -= 1
                        if ind[ni][nj] == 0:
                            q.append((ni, nj))
        return max_path



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