<p>Given an <code>m x n</code> integer matrix <code>heightMap</code> representing the height of each unit cell in a 2D elevation map, return <em>the volume of water it can trap after raining</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/04/08/trap1-3d.jpg" style="width: 361px; height: 321px;" />
<pre>
<strong>Input:</strong> heightMap = [[1,4,3,1,3,2],[3,2,1,3,2,4],[2,3,3,2,3,1]]
<strong>Output:</strong> 4
<strong>Explanation:</strong> After the rain, water is trapped between the blocks.
We have two small ponds 1 and 3 units trapped.
The total volume of water trapped is 4.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/04/08/trap2-3d.jpg" style="width: 401px; height: 321px;" />
<pre>
<strong>Input:</strong> heightMap = [[3,3,3,3,3],[3,2,2,2,3],[3,2,1,2,3],[3,2,2,2,3],[3,3,3,3,3]]
<strong>Output:</strong> 10
</pre>

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

<ul>
	<li><code>m == heightMap.length</code></li>
	<li><code>n == heightMap[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 200</code></li>
	<li><code>0 &lt;= heightMap[i][j] &lt;= 2 * 10<sup>4</sup></code></li>
</ul>

class Solution:
    def trapRainWater(self, heightMap: List[List[int]]) -> int:
        from queue import PriorityQueue

        if len(heightMap) == 0:
            return 0
        elif len(heightMap[0]) == 0:
            return 0
        n, m = len(heightMap), len(heightMap[0])
        if n==1 and m==1:
            return 0
        
        q = PriorityQueue()
        visited = set()
        directions = [(1, 0), (-1, 0), (0, 1), (0, -1)]

        for i in range(n-1):
            q.put((heightMap[i][0], (i, 0)))
            visited.add((i, 0))
            q.put((heightMap[i+1][m-1], (i+1, m-1)))
            visited.add((i+1, m-1))
        for j in range(m-1):
            q.put((heightMap[0][j+1], (0, j+1)))
            visited.add((0, j+1))
            q.put((heightMap[n-1][j], (n-1, j)))
            visited.add((n-1, j))

        ans = 0
        
        while not q.empty():
            v, (i, j) = q.get()
            for di, dj in directions:
                ni, nj = i+di, j+dj
                if (ni >= 0) and (ni < n) and (nj >= 0) and (nj < m):
                    if (ni, nj) not in visited:
                        q.put((max(v, heightMap[ni][nj]), (ni, nj)))
                        visited.add((ni, nj))
                        if v >= heightMap[ni][nj]:
                            ans += v - heightMap[ni][nj]
                        # escape[ni][nj] = max(v, heightMap[ni][nj])


        # for i in range(n):
        #     for j in range(m):
        #         if escape[i][j] > heightMap[i][j]:
        #             ans += escape[i][j] - heightMap[i][j]

        return ans

        return ans
23000