<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:
    import heapq
    dirs = [[-1,0],[1,0],[0,1],[0,-1]]

    def trapRainWater(self, heightMap: List[List[int]]) -> int:
        if not heightMap or not heightMap[0]:
            return 0
        m, n = len(heightMap), len(heightMap[0])
        pq = []
        visited = {}

        for i in range(m):
            c1 = (heightMap[i][0], i ,0)
            c2 = (heightMap[i][n-1], i, n-1)
            heapq.heappush(pq, c1)
            heapq.heappush(pq, c2)
            visited[(i,0)] = 1
            visited[(i, n-1)] = 1 
        
        for j in range(n):
            c1 = (heightMap[0][j], 0, j )
            c2 = ( heightMap[m-1][j], m-1, j)
            heapq.heappush(pq, c1)
            heapq.heappush(pq, c2)
            visited[(0,j)] = 1
            visited[(m-1,j)] = 1        
        
        res = 0
        while pq:
            cell = heapq.heappop(pq)
            
            for direction in self.dirs:
                x = cell[1]+direction[0]
                y = cell[2]+direction[1]
                if x<0 or y<0 or x>=m or y>=n or (x,y) in visited:
                    continue
                visited[(x,y)] = 1

                if heightMap[x][y] < cell[0]:
                    res += cell[0]-heightMap[x][y]
                
                nc = (max(cell[0],heightMap[x][y]), x,y)
                heapq.heappush(pq, nc)

        return res
22400