<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}};
        
        
        // <height, row, col>
        priority_queue< pair<int,pair<int,int>> , vector<pair<int, pair<int,int>>>, greater<pair<int, pair<int,int>>>>min_heap;
        
        // add all boundary cells
        // top and bottom cells
        for(int i = 0; i < N; i++) {
            min_heap.push({heights[0][i],{ 0, i}}),
            min_heap.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++) {
            min_heap.push({heights[i][0], {i, 0}}),
            min_heap.push({heights[i][N-1], {i, N-1}});
            visited[i][0] = true, visited[i][N-1] = true;
        }
            
        while(!min_heap.empty()) {
            int height= min_heap.top().first;
            int row = min_heap.top().second.first;
            int col = min_heap.top().second.second;
            min_heap.pop();
            
            // explore the neighbouring cells
            for(auto dir: directions) {
                int r = row + dir[0], c = col + dir[1];
                // process the 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];
                    
                    min_heap.push({max(height, heights[r][c]), {r, c}});
                }
            }
        }
        return vol;
    }
};
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