<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:
        m, n = len(heightMap), len(heightMap[0])
        directions = ((-1, 0), (1, 0), (0, -1), (0, 1))
        pq = []
        seen = [[False for _ in range(n)] for __ in range(m)]
        for row in range(m):
            pq.append((heightMap[row][0], row, 0))
            seen[row][0] = True
            if n > 1:
                pq.append((heightMap[row][n - 1], row, n - 1))
                seen[row][n - 1] = True
        for col in range(1, n - 1):
            pq.append((heightMap[0][col], 0, col))
            seen[0][col] = True
            if m > 1:
                pq.append((heightMap[m - 1][col], m - 1, col))
                seen[m - 1][col] = True
        heapify(pq)
        volume = 0
        while pq:
            boundary, row, col = heappop(pq)
            for dr, dc in directions:
                r, c = row + dr, col + dc
                if 0 <= r < m and 0 <= c < n and not seen[r][c]:
                    volume += max(0, boundary - heightMap[r][c])
                    heightMap[r][c] = max(heightMap[r][c], boundary)
                    heappush(pq, (heightMap[r][c], r, c))
                    seen[r][c] = True
        return volume

18600