<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):
        heap = []
        L = len(heightMap)
        W = len(heightMap[0])
        visited = set()
        for i in range(L):
            for j in range(W):
                if i in (0, L-1) or j in (0, W-1):
                    heapq.heappush(heap, (heightMap[i][j], i, j))
                    visited.add((i, j))
        

        result = 0
        max_water = 0
        while heap:
            cur_water, cur_i, cur_j = heapq.heappop(heap)
            max_water = max(max_water, cur_water)

            for dir_i, dir_j in [(-1, 0), (1, 0), (0, 1), (0, -1)]:
                nxt_i, nxt_j = cur_i + dir_i, cur_j + dir_j
                if 0<=nxt_i<L and 0<=nxt_j<W and (nxt_i, nxt_j) not in visited:
                    visited.add((nxt_i, nxt_j))
                    result += max(0, max_water - heightMap[nxt_i][nxt_j])
                    heapq.heappush(heap, (heightMap[nxt_i][nxt_j], nxt_i, nxt_j))
      #  print(heap)
        return result
23500