<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 ppi pair<int, pair<int, int>> 
class Solution {
public:
    int trapRainWater(vector<vector<int>>& height) {
        int n = size(height), m = size(height[0]);
        // basically we will keep a track that, 
        // what is the first minimum height > height[i][j]
        // with which we are visiting it......

        priority_queue<ppi, vector<ppi>, greater<>> pq;

        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({height[i][j], {i, j}});
            }
        }

        int ans = 0;
        int dx[4] = {-1, 0, 0, 1};
        int dy[4] = {0, 1, -1, 0};
        vector<vector<int>> vis(n+1, vector<int>(m+1, 0));
        while(!pq.empty()) {
            int i = pq.top().second.first;
            int j = pq.top().second.second;
            int h = pq.top().first;
            pq.pop();

            if (vis[i][j]) continue;
            vis[i][j] = 1;
            ans += (h - height[i][j]);
            for (int k=0; k<4; k++) {
                int r = i + dx[k], c = j + dy[k];
                if (r>=0 && c>=0 && r<n && c<m) {
                    pq.push({max(height[r][c], h), {r, c}});
                }
            }
        }
        return ans;
    }
};
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