<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, H: List[List[int]]) -> int:
        dirs = [(-1,0), (1,0), (0,1), (0,-1)]
        m, n = len(H), len(H[0])
        S = []
        d = [[float('inf') for _ in range(n)] for _ in range(m)]
        for i in range(m):
            for j in range(n):
                if i==0 or j==0 or i==(m-1) or j==(n-1):
                    S.append((H[i][j], (i, j)))
                    d[i][j] = H[i][j]
        heapq.heapify(S)
        while len(S):
            dist, u = heapq.heappop(S)
            for dx, dy in dirs:
                v = (u[0]+dx, u[1]+dy)
                if 0<=v[0]<m and 0<=v[1]<n and d[v[0]][v[1]]>max(dist, H[v[0]][v[1]]):
                    d[v[0]][v[1]] = max(dist, H[v[0]][v[1]])
                    heapq.heappush(S, (d[v[0]][v[1]], v))
        water = 0
        for i in range(m):
            for j in range(n):
                water += max(d[i][j]-H[i][j], 0)
        return water
19400