Trapping Rain Water II
<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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0400-0499/0407.Trapping%20Rain%20Water%20II/images/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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0400-0499/0407.Trapping%20Rain%20Water%20II/images/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>
Breadth-First Search; Array; Matrix; Heap (Priority Queue)
class Solution:
    def trapRainWater(self, heightMap: List[List[int]]) -> int:
        m, n = len(heightMap), len(heightMap[0])
        vis = [[False] * n for _ in range(m)]
        pq = []
        for i in range(m):
            for j in range(n):
                if i == 0 or i == m - 1 or j == 0 or j == n - 1:
                    heappush(pq, (heightMap[i][j], i, j))
                    vis[i][j] = True
        ans = 0
        dirs = (-1, 0, 1, 0, -1)
        while pq:
            h, i, j = heappop(pq)
            for a, b in pairwise(dirs):
                x, y = i + a, j + b
                if x >= 0 and x < m and y >= 0 and y < n and not vis[x][y]:
                    ans += max(0, h - heightMap[x][y])
                    vis[x][y] = True
                    heappush(pq, (max(h, heightMap[x][y]), x, y))
        return ans
