<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:
        if not heightMap or not heightMap[0]:
            return 0
        
        m, n = len(heightMap), len(heightMap[0])
        heap = []
        visited = [[False] * n for _ in range(m)]
        
        # Add all border cells to the heap
        for i in range(m):
            for j in range(n):
                if i == 0 or i == m-1 or j == 0 or j == n-1:
                    heappush(heap, (heightMap[i][j], i, j))
                    visited[i][j] = True
        
        directions = [(0, 1), (0, -1), (1, 0), (-1, 0)]
        water = 0
        max_height = 0
        
        while heap:
            height, i, j = heappop(heap)
            max_height = max(max_height, height)
            
            for di, dj in directions:
                ni, nj = i + di, j + dj
                if 0 <= ni < m and 0 <= nj < n and not visited[ni][nj]:
                    visited[ni][nj] = True
                    if heightMap[ni][nj] < max_height:
                        water += max_height - heightMap[ni][nj]
                    heappush(heap, (heightMap[ni][nj], ni, nj))
        
        return water

18800