<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  pr pair<int,pair<int, int>>
class Solution {
public:
    int dy[4] = {1,-1,0,0};
    int dx[4] = {0,0,1,-1};
    int trapRainWater(vector<vector<int>>& heightMap) {
        int m = heightMap.size();
        int n = heightMap[0].size();
        vector<vector<int>> vis(m, vector<int>(n,0));
        priority_queue<pr, vector<pr>, greater<pr>> pq;

        for(int i=0; i<m; i++){
            for(int j=0; j<n; j++){
                //first checking the boundary
                if(i==0 or j==0 or i==m-1 or j==n-1){
                    vis[i][j] = true;
                    pq.push({heightMap[i][j],{i,j}});
                }
            }
        }
        int water = 0;
        while(!pq.empty()){
            auto front = pq.top();
            pq.pop();
            int x = front.second.first;
            int y = front.second.second;
            for(int i=0; i<4; i++){
                int xx = x + dx[i];
                int yy = y + dy[i];
                if(xx>=0 && yy>=0 && xx<m && yy<n && !vis[xx][yy]){
                    int gap = max(0, front.first-heightMap[xx][yy]);
                    water += gap;
                    pq.push({heightMap[xx][yy] + gap, {xx,yy}});
                    vis[xx][yy] = true;
                    }
                }
            }
        return water;
    }
};
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