<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:
        
        n, m = len(heightMap), len(heightMap[0])
        h = []
        queued = [[0 for c in range(m)] for r in range(n)]
        seen = [[0 for c in range(m)] for r in range(n)]
        mx = 0
        total = 0
        
        def enqueue(i, j):
            if not queued[i][j] and not seen[i][j]:
                heapq.heappush(h, [heightMap[i][j], i, j])
                queued[i][j] = 1
                
        def enqueue_neighbors(i, j):
            if i >= n or j >= m:
                return []
            if i != 0:
                enqueue(i-1, j)
            if i != n-1:
                enqueue(i+1, j)
            if j != 0:
                enqueue(i, j-1)
            if j != m-1:
                enqueue(i, j+1)

        for j in range(m):
            enqueue(0, j)
            enqueue(n-1, j)
        for i in range(n):
            enqueue(i, 0)
            enqueue(i, m-1)
        
        while h:
            val, i, j = heapq.heappop(h)
            if val < mx:
                total += mx - val
            mx = max(mx, val)
            seen[i][j] = 1
            enqueue_neighbors(i, j)
        
        return total
19300