<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>>& mat)
    {
        int n=mat.size();
        int m=mat[0].size();
        priority_queue<pair<int,pair<int,int>>,vector<pair<int,pair<int,int>>>,greater<pair<int,pair<int,int>>>>pq;
        vector<vector<bool>>vis(n,vector<bool>(m,false));
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<m;j++)
            {
                if(i==0||j==0||i==n-1||j==m-1)
                {
                    pq.push({mat[i][j],{i,j}});
                    vis[i][j]=true;
                }
            }
        }
        vector<pair<int,int>>d={{-1,0},{1,0},{0,-1},{0,1}};
    int ans=0;
        while(pq.size())
        {
            auto temp=pq.top();
            pq.pop();
            int val=temp.first;
            int x=temp.second.first;
            int y=temp.second.second;
            for(auto it:d)
            {
                int nx=x+it.first;
                int ny=y+it.second;
                if(nx>=0&&ny>=0&&nx<n&&ny<m&&vis[nx][ny]==false)
                {
                    vis[nx][ny]=true;
                    ans+=max(0,val-mat[nx][ny]);
                    pq.push({max(val,mat[nx][ny]),{nx,ny}});
                }
            }
        }
        return ans;
    }
};
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