<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:
        rows, cols = len(heightMap), len(heightMap[0])
        heap = []
        directions = [(1, 0), (0, 1), (-1, 0), (0, -1)]
        res = 0

        for r in range(rows): 
            for c in range(cols): 
                if r in {0, rows - 1} or c in {0, cols - 1}: 
                    heapq.heappush(heap, (heightMap[r][c], r, c))
                    heightMap[r][c] = -1
        
        while heap: 
            h, r, c = heapq.heappop(heap)
            for dr, dc in directions: 
                x, y = r + dr, c + dc
                if x not in range(rows) or y not in range(cols) or heightMap[x][y] == -1: 
                    continue
                
                res += max(0, h - heightMap[x][y])
                heapq.heappush(heap, (max(h, heightMap[x][y]), x, y))
                heightMap[x][y] = -1
        return res
            




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