<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>>& heightMap) {
        int m = heightMap.size(), n = heightMap[0].size();
        int res = 0, maxH = -1;
        unordered_set<int> visited;
        priority_queue<vector<int>, vector<vector<int>>, greater<vector<int>>> myq;
        vector<vector<int>> dirs{{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
        for(int i = 0; i < m; ++i)
        for(int j = 0; j < n; ++j)
        {
            if(i == 0 || i == m - 1 || j == 0 || j == n - 1)
            {
                int index = i * n + j;
                myq.push({heightMap[i][j], index});
                
            }    
        }
        
        while(!myq.empty())
        {
            auto v = myq.top();
            myq.pop();

            maxH = max(v[0], maxH);
            int i = v[1] / n, j = v[1] % n;
            for(auto &d : dirs)
            {
                int x = i + d[0], y = j + d[1];
                int index = x * n + y;
                if(x <= 0 || y <= 0 || x >= m-1 || y >= n-1
                   || visited.count(index))
                    continue;
                res += (heightMap[x][y] < maxH) ? maxH - heightMap[x][y] : 0;
                myq.push({heightMap[x][y], index});
                visited.insert(index);
            }
        }
        return res;
    }
};
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