<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 = len(heightMap)
        n = len(heightMap[0])
        res = 0
        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:
                    heap.append((heightMap[i][j], (i, j)))
                    heightMap[i][j] = -1
        rows = [-1, 0, 1, 0]
        cols = [0, 1, 0, -1]
        heapq.heapify(heap)
        level = 0
        res = 0
        while heap:
            height, (i, j) = heapq.heappop(heap)
            if height < level:
                res += level - height
            level = max(level, height)
            for d in range(4):
                r = i + rows[d]
                c = j + cols[d]
                if 0 <= r < m and 0 <= c < n and heightMap[r][c] > -1:
                    heapq.heappush(heap, (heightMap[r][c], (r, c)))
                    heightMap[r][c] = -1
        return res

        
19200