<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>

import heapq

DIRECTIONS = [(1, 0), (-1, 0), (0, 1), (0, -1)]

class Solution:
    def trapRainWater(self, heightMap: List[List[int]]) -> int:
        if len(heightMap) == 0 or len(heightMap[0]) == 0:
            return 0
        
        m = len(heightMap)
        n = len(heightMap[0])
        res = 0
        heap_arr = []
        visited = set([])
        
        # starts from boundary
        for i in range(m):
            for j in range(n):
                if (i == 0 or j == 0 or i == m-1 or j == n-1):
                    heapq.heappush(heap_arr, (heightMap[i][j], i, j))
                    visited.add((i, j))
        
        # expand from lowest of the boundary
        while len(heap_arr) != 0:
            top_val, top_x, top_y = heapq.heappop(heap_arr)
            for x_step, y_step in DIRECTIONS:
                next_x, next_y = top_x + x_step, top_y + y_step
                if 0 <= next_x < m and 0 <= next_y < n and (next_x, next_y) not in visited:
                    # (top.x, top.y) is P
                    # P's water line is at top.val
                    # (nx, ny) is Q
                    # Q's height is heights[nx][ny]
                    next_h = max(top_val, heightMap[next_x][next_y])
                    next_water = next_h - heightMap[next_x][next_y]
                    res += next_water
                    heapq.heappush(heap_arr, (next_h, next_x, next_y))
                    visited.add((next_x, next_y))
        return res
                
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