<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>>& height) {
        priority_queue<pair<int,pair<int,int>>, vector<pair<int,pair<int,int>>>,greater<>>pq;

        //boundary

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

        vector<vector<bool>> visited(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({height[i][j],{i,j}});
                    visited[i][j] = true;
                }
            }
        }
        vector<vector<int>>dir= {{-1,0},{1,0},{0,-1},{0,1}};
        int trapwater = 0;
        while(pq.size()>0){
            auto top = pq.top();
            pq.pop();
            int h = top.first;
            int r = top.second.first;
            int c = top.second.second;

            for(int i = 0;i < dir.size();i++){
                int nr = dir[i][0]+r;
                int nc =dir[i][1]+c;
                if(nr >= 0 && nc >= 0 && nr < n && nc < m && visited[nr][nc] == false){
                    visited[nr][nc] = true;
                    trapwater += max(0,h-height[nr][nc]);
                    pq.push({max(h,height[nr][nc]),{nr,nc}});
                }
            }
        }
        return trapwater;
    }
};
17006