<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:
// TC O(NlogN), N is total count of heights, SC O(N)
    int trapRainWater(vector<vector<int>>& heightMap) {
        int m = heightMap.size(), n = heightMap[0].size();
        unordered_set<int> visited;

        priority_queue<vector<int>, vector<vector<int>>, greater<vector<int>>> pq;

        for(int i = 0; i < m; ++i)
        for(int j = 0; j < n; ++j) {
            if(i == 0 || j == 0 || i == m-1 || j == n-1) {
                int index = i*n+j;
                visited.insert(index);
                pq.push({heightMap[i][j], index});
            }
        }

        vector<vector<int>> dirs = {{1,0}, {0,1}, {-1,0}, {0,-1}};
        int res = 0, curMaxH = 0;
        while(!pq.empty()) {
            auto vec = pq.top();
            pq.pop();

            curMaxH = max(curMaxH, vec[0]);
            int index = vec[1];
            int i = index / n, j = index % n;

            for(auto &dir : dirs) {
                int x = i + dir[0], y = j + dir[1], tmp = x*n+y;
                if(x < 0 || y < 0 || x == m || y == n || visited.count(tmp))
                    continue;
                
                if(heightMap[x][y] < curMaxH)
                    res += curMaxH-heightMap[x][y];
                pq.push({heightMap[x][y], tmp});
                visited.insert(tmp);
            }
        }

        return res;
    }
};
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