<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>>& h) {
        ios_base :: sync_with_stdio( false );
        
        priority_queue< vector<int> , vector<vector<int>> , greater<vector<int>> >pq;
        int cols = h[0].size() , rows = h.size();
        int result = 0;

        vector< vector<int>> vis(rows , vector<int>(cols , 0));
        for(int i=0 ; i<rows ; i++){
            for(int j=0 ; j<cols ; j++){
                if( i==0 || j==0 || i== rows-1 || j== cols-1){
                    pq.push( { h[i][j] , i , j } );
                    vis[i][j] = true;
                }
            }
        }

        int up[]   = { 1, -1 , 0 , 0};
        int down[] = { 0,  0 , 1 ,-1};
        int minHeight = 0;

        while( !pq.empty() ){
            int mini = pq.top()[0];
            int idx = pq.top()[1];
            int jdx = pq.top()[2];
            minHeight = max(minHeight , mini);
            pq.pop();

            for(int i=0 ; i<4 ; i++){
                int tempi = idx + up[i];
                int tempj = jdx + down[i];
                if(tempi >=0 && tempi < rows && tempj >=0 && tempj < cols && vis[tempi][tempj] == false){
                    pq.push( { h[tempi][tempj] , tempi , tempj } );
                    vis[tempi][tempj] = true;
                    result += (minHeight > h[tempi][tempj])?  (minHeight - h[tempi][tempj]) : 0 ;
                }
            }
        }
        return result;
    }
};


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