<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])
        if m == 1 or n == 1:
            return 0
        height = [[False for i in range(n)] for j in range(m)]

        ans = 0
        heap = []
        for i in range(m):
            heapq.heappush(heap, (heightMap[i][0], i, 0))
            heapq.heappush(heap, (heightMap[i][n - 1], i, n - 1))
        for i in range(1, n - 1):
            heapq.heappush(heap, (height[0][i], 0, i))
            heapq.heappush(heap, (height[m - 1][i], m - 1, i))
        
        while heap:
            water, i, j = heapq.heappop(heap)
            if height[i][j]:
                continue
            height[i][j] = True
            ans += max(0, water - heightMap[i][j])
            for dx, dy in [(-1, 0), (1, 0), (0, -1), (0, 1)]:
                if 0 <= dx + i < m and 0 <= dy + j < n and not height[dx + i][dy + j]:
                    curMax = max(heightMap[i][j], water)
                    heapq.heappush(heap, (curMax, dx + i, dy + j))
        return ans
19100