<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>

import heapq

class Solution:
    def trapRainWater(self, heightMap: List[List[int]]) -> int:
        if not heightMap or not heightMap[0]:
            return 0
        
        m, n = len(heightMap), len(heightMap[0])
        # Min-heap priority queue
        min_heap = []
        # Visited matrix to keep track of visited cells
        visited = [[False] * n for _ in range(m)]
        
        # Add all boundary cells to the heap and mark them as visited
        for i in range(m):
            for j in [0, n - 1]:
                heapq.heappush(min_heap, (heightMap[i][j], i, j))
                visited[i][j] = True
        
        for j in range(n):
            for i in [0, m - 1]:
                if not visited[i][j]:
                    heapq.heappush(min_heap, (heightMap[i][j], i, j))
                    visited[i][j] = True
        
        # Directions for moving up, down, left, right
        directions = [(0, 1), (1, 0), (0, -1), (-1, 0)]
        trapped_water = 0
        
        while min_heap:
            height, x, y = heapq.heappop(min_heap)
            
            for dx, dy in directions:
                nx, ny = x + dx, y + dy
                
                if 0 <= nx < m and 0 <= ny < n and not visited[nx][ny]:
                    visited[nx][ny] = True
                    trapped_water += max(0, height - heightMap[nx][ny])
                    heapq.heappush(min_heap, (max(height, heightMap[nx][ny]), nx, ny))
        
        return trapped_water

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