<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>

# define p pair<int,pair<int,int>> 

class Solution {
public:

// first saare corner walo ko min heap me dalo
//phir bfs start kro aur tot calculate kro

    int trapRainWater(vector<vector<int>>& h) {

        int m=h.size();
        int n=h[0].size();

        vector<vector<bool>> v(m,vector<bool>(n,false));

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

        for(int i=0;i<m;i++)
        {
            pq.push({h[i][0],{i,0}});
            pq.push({h[i][n-1],{i,n-1}});

            v[i][0]= v[i][n-1]=true;
                        
        }

        for(int j=0;j<n;j++)
        {
            pq.push({h[0][j],{0,j}});
            pq.push({h[m-1][j],{m-1,j}});

            v[0][j]= v[m-1][j]=true;
                        
        }

        int tot=0;

        vector<vector<int>> dir={{-1,0}, {1,0}, {0,1}, {0,-1}}; 

        while(!pq.empty())
        {
            auto a=pq.top();
            pq.pop();

            int hei = a.first;
            int x=a.second.first;
            int y=a.second.second;

            for(auto i : dir )
            {
                int nx=x+i[0];
                int ny= y + i[1];

                if(nx>=0 && ny>=0 && nx < m && ny < n && !v[nx][ny])
                {
                    tot+=max(0,hei-h[nx][ny]);
                    v[nx][ny]=true;

                    pq.push({max(hei,h[nx][ny]), {nx,ny}});
                }
            }
        }

        return tot;


        
    }
};
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