<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) {
        
        priority_queue<vector<int>,vector<vector<int>>,greater<vector<int>>> pq;
        
        int m = heightMap.size();
        int n = heightMap[0].size();
        
        vector<vector<bool>> v(m,vector<bool>(n,0));
        
        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){
                    pq.push({heightMap[i][j],i,j});
                    v[i][j]=1;
                }
            }
        }
        
        int minBH = 0; 
        int ans = 0;
        while(!pq.empty()){
        
            int currHeight = pq.top()[0];
            int x = pq.top()[1];
            int y = pq.top()[2];
            //cout<<x<<" "<<y<<endl;
            pq.pop();
            
            minBH = max(minBH, currHeight);
                        
            int dir[4][2] = {{1,0},{0,1},{0,-1},{-1,0}};
            
            for(auto d: dir){
                int newX = d[0]+x;
                int newY = d[1]+y;
                
                if(newX >= 0 && newX < m && newY >= 0 && newY < n && !v[newX][newY]){
                    
                    v[newX][newY] = 1;

                    pq.push({heightMap[newX][newY], newX, newY});
                    
                    if(heightMap[newX][newY] < minBH) ans+= (minBH-heightMap[newX][newY]);
                    
                }
            }
        }
        return ans;
    }
};
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