<p>You are given an <code>m x n</code> integer matrix <code>grid</code>, where you can move from a cell to any adjacent cell in all <code>4</code> directions.</p>

<p>Return <em>the number of <strong>strictly</strong> <strong>increasing</strong> paths in the grid such that you can start from <strong>any</strong> cell and end at <strong>any</strong> cell. </em>Since the answer may be very large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

<p>Two paths are considered different if they do not have exactly the same sequence of visited cells.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/05/10/griddrawio-4.png" style="width: 181px; height: 121px;" />
<pre>
<strong>Input:</strong> grid = [[1,1],[3,4]]
<strong>Output:</strong> 8
<strong>Explanation:</strong> The strictly increasing paths are:
- Paths with length 1: [1], [1], [3], [4].
- Paths with length 2: [1 -&gt; 3], [1 -&gt; 4], [3 -&gt; 4].
- Paths with length 3: [1 -&gt; 3 -&gt; 4].
The total number of paths is 4 + 3 + 1 = 8.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> grid = [[1],[2]]
<strong>Output:</strong> 3
<strong>Explanation:</strong> The strictly increasing paths are:
- Paths with length 1: [1], [2].
- Paths with length 2: [1 -&gt; 2].
The total number of paths is 2 + 1 = 3.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 1000</code></li>
	<li><code>1 &lt;= m * n &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= grid[i][j] &lt;= 10<sup>5</sup></code></li>
</ul>

"""

___
Goal:
___


return # of strictly increasing paths in grid

___
Examples:
___

Input: grid = [[1,1],[3,4]]
Output: 8

- Paths with length 1: [1], [1], [3], [4].
- Paths with length 2: [1 -> 3], [1 -> 4], [3 -> 4].
- Paths with length 3: [1 -> 3 -> 4].

Input: grid = [[1],[2]]
Output: 3

- Paths with length 1: [1], [2]
- Paths with length 2: [1, 2]

___
Assumptions:
___


1. must start from any cell and end at any cell

2. 1 <= len(grid), len(grid[0]) <= 1000

3. 1 <= m * n <= 10_000

4. 1 <= grid[i][j] <= 10_000

5. can traverse from N, S, W, E direction on any cell

___
Observations:
___

1. Naive method could be N!

2. recurrence relationship:
    - count(parent) = 1 + [count(child) for child in directions if child strictly_increasing to parent]

[1]
[1 -> 3]
[1 -> 3 -> 4]

Topological Sorting:
    - efficiently allows us prioritize traversing leaf nodes
    - could iterate over grid:
        - indegree dictated by how many neighbors smaller than you
            ex. 4 has indegree of 2 versus 1 having indegree of 0  
"""


class Solution:
    def countPaths(self, grid: List[List[int]]) -> int:
        
        rows, cols = len(grid), len(grid[0])

        indegree = {(i, j):0 for i in range(rows) for j in range(cols)}
        directions = [(0, 1), (0, -1), (1, 0), (-1, 0)]
        parent = {(i, j):[] for i in range(rows) for j in range(cols)}
        memo = {(i, j):1 for i in range(rows) for j in range(cols)}
        
        for x in range(rows):
            for y in range(cols):
                for dx, dy in directions:
                    if 0 <= (a := x + dx) < rows and 0 <= (b := y + dy) < cols and grid[a][b] > grid[x][y]:
                        parent[(x, y)].append((a, b))
                        indegree[(a, b)] += 1
        

        queue = deque([node for node, d in indegree.items() if not d])
        

        while queue:
            node = queue.popleft()
            
            for p in parent[node]:
                memo[p] += memo[node]
                indegree[p] -= 1
                if not indegree[p]:
                    queue.append(p)


        return sum(memo.values()) % (10 ** 9 + 7)
121199