<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>>& a) {
        int n=a.size();
        int m=a[0].size();
        priority_queue<pair<int,pair<int,int>>,vector<pair<int,pair<int,int>>>,greater<pair<int,pair<int,int>>>>pq;
        vector<vector<int>>vis(n,vector<int>(m,0));
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(i==0||i==n-1||j==0||j==m-1){pq.push({a[i][j],{i,j}});
                vis[i][j]=1;}
            }
        }
        int l=0,res=0;
        while(!pq.empty()){
            auto p=pq.top();
            pq.pop();
            
            int ht=p.first;
            int i=p.second.first;
            int j=p.second.second;
            l=max(l,ht);
            vector<int>r={-1,0,1,0};
            vector<int>c={0,1,0,-1};
            
                for(int y=0;y<4;y++){
                    int nr=i+r[y];
                    int nc=j+c[y];
                    if(nr>=0&&nr<n&&nc>=0&&nc<m&&vis[nr][nc]==0){
                        pq.push({a[nr][nc],{nr,nc}});
                        //cout<<a[nr][nc]<<" "<<l<<" ";
                        res+=max(0,l-a[nr][nc]);
                        vis[nr][nc]=1;
                    }
                }
            

        }
        return res;

    }
};
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