<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:
        m, n = len(heightMap), len(heightMap[0])
        if m < 3 or n < 3: return 0
        # to simplify the code
        def adjacent(i,j):
            return [(i-1,j), (i+1,j), (i,j-1), (i,j+1)]
        
        # first we sort all heights from the matrix and store their corresponding positions
        # heights will be [(h1,[p1, p2,...]), (h2,[p1, p2, ...]), ...]
        d = defaultdict(list)
        for i in range(m):
            for j in range(n):
                d[heightMap[i][j]].append((i,j))
        heights = sorted(d.items())
        # initialization
        volumn, area, h_lower = 0, 0, heights[0][0]
        level = [[1 for j in range(n)] for i in range(m)]
        for h, positions in heights:
            volumn += area*(h-h_lower)
            h_lower = h
            leak = []
            for i, j in positions:
                # due to height rising, now this position may hold water
                level[i][j] = 0
                area += 1
                if i == 0 or i == m-1 or j == 0 or j == n-1 or any([level[a][b] == -1 for a, b in adjacent(i,j)]):
                    # this position is reachable from outside, therefore cannot hold water
                    leak.append((i,j))
                    level[i][j] = -1
                    area -= 1
            while leak:
                i, j = leak.pop()
                for a, b in adjacent(i,j):
                    if 0 <= a < m and 0 <= b < n and not level[a][b]:
                        # new leaking position found through DFS
                        leak.append((a,b))
                        level[a][b] = -1
                        area -= 1

        return volumn
21600