<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:
        nr = len(heightMap)
        if nr <= 1:
            return 0
        nc = len(heightMap[0])
        min_heap, visited = [], defaultdict(bool)
        for i in range(nr):
            for j in range(nc):
                if i == 0 or i == nr - 1 or j == 0 or j == nc - 1:
                    heapq.heappush(min_heap, (heightMap[i][j], i, j))
                    visited[(i, j)] = True
        
        res = 0
        while min_heap:
            height, x, y = heapq.heappop(min_heap)
            horizon = height
            for d in [[-1, 0], [1, 0], [0, -1], [0, 1]]:
                nx, ny = x + d[0], y + d[1]
                if 0 <= nx < nr and 0 <= ny < nc and not visited[(nx, ny)]:
                    visited[(nx, ny)] = True
                    if heightMap[nx][ny] < height: 
                        res += height - heightMap[nx][ny]
                    heapq.heappush(min_heap, (max(height, heightMap[nx][ny]), nx, ny))
        return res
        

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