<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])
        visited = set()
        heap = []

        # Add boundary cells to heap and mark them as visited
        for i in range(m):
            for j in [0, n - 1]:
                heapq.heappush(heap, (heightMap[i][j], i, j))
                visited.add((i, j))
        for j in range(1, n - 1):
            for i in [0, m - 1]:
                heapq.heappush(heap, (heightMap[i][j], i, j))
                visited.add((i, j))

        total_water = 0
        directions = [(-1, 0), (1, 0), (0, -1), (0, 1)]  # Up, Down, Left, Right

        while heap:
            height, x, y = heapq.heappop(heap)
            for dx, dy in directions:
                nx, ny = x + dx, y + dy
                # Check bounds and if the cell has not been visited
                if 0 <= nx < m and 0 <= ny < n and (nx, ny) not in visited:
                    visited.add((nx, ny))
                    neighbor_height = heightMap[nx][ny]
                    trapped = max(0, height - neighbor_height)
                    total_water += trapped
                    # Update neighbor's height to the max between its own height and the current water level
                    heapq.heappush(heap, (max(height, neighbor_height), nx, ny))
        return total_water

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