<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 {
public:
    int trapRainWater(vector<vector<int>>& heights) {
        int vol = 0;
        const int M = heights.size(), N = heights[0].size();
        vector<vector<bool>> visited(M, vector<bool>(N, false));
        vector<vector<int>> directions = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
            
        auto comp = [&](const array<int, 3>& a, const array<int, 3>& b) { return a[0] >= b[0]; };
        
        // <height, row, col>
        priority_queue<array<int, 3>, vector<array<int, 3>>, decltype(comp)> pq(comp);
        
        // add all boundary cells
        // top and bottom cells
        for (int i=0; i<N; ++i) {
            pq.push({heights[0][i], 0, i});
            pq.push({heights[M-1][i], M-1, i});
            visited[0][i] = true;
            visited[M-1][i] = true;
        }
        
        // right and left cells
        for (int i=0; i<M; ++i) {
            pq.push({heights[i][0], i, 0});
            pq.push({heights[i][N-1], i, N-1});
            visited[i][0] = true;
            visited[i][N-1] = true;
        }
        
        while (!pq.empty()) {
            auto [height, row, col] = pq.top();
            pq.pop();
            
            // explore the neighbouring cells
            for (auto dir : directions) {
                int r = row + dir[0], c = col + dir[1];
                // process in inner cell
                if (r>=0 && r<M && c>=0 && c < N && !visited[r][c]) {
                    // mark current cell as visited
                    visited[r][c] = true;
                    // if current is smaller, it can store water
                    if (heights[r][c] < height)
                        vol += height - heights[r][c];
                    
                    pq.push({max(height, heights[r][c]), r, c});
                }
            }
        }
        return vol;
    }
};
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