<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) {

        set<pair<int, pair<int,int>>>s;
        vector<vector<int>> vis(h.size(), vector<int> (h[0].size(), 0));
        for(int i = 0; i<h.size(); i++)
        {
            vis[i][0] = 1;
            vis[i][h[0].size()-1] = 1;
            s.insert({h[i][0], {i, 0}});
            s.insert({h[i][h[0].size()-1], {i, h[0].size()-1}});
        }
        for(int j = 0; j<h[0].size(); j++)
        {
            vis[0][j] = 1;
            vis[h.size()-1][j] = 1;
            s.insert({h[0][j], {0, j}});
            s.insert({h[h.size()-1][j], {h.size()-1,j}});
        }
        vector<pair<int,int>> nbhr = {{1,0}, {-1,0}, {0,1}, {0,-1}};
        int ans = 0;
        while(!s.empty())
        {
            auto x = *s.begin();
            s.erase(s.begin());
            int i = x.second.first;
            int j = x.second.second;
            for(auto it: nbhr)
            {
                int ip = i + it.first;
                int jp = j + it.second;
                if(ip >= 0 && ip < h.size() && jp >= 0 && jp < h[0].size())
                {
                    if(vis[ip][jp] == 0)
                    {
                        if(h[ip][jp] < h[i][j])
                        {
                            ans += h[i][j] - h[ip][jp];
                            h[ip][jp] = h[i][j];
                            vis[ip][jp] = 1;
                            s.insert({h[ip][jp], {ip, jp}});
                        }
                        else
                        {
                            vis[ip][jp] = 1;
                            s.insert({h[ip][jp], {ip, jp}});
                        }
                    }
                }
            }
        }
        return ans;
    }
};
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