<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>>& heights) {
        int n = heights.size();
        int m = heights[0].size();

        vector<vector<bool>> visited(n, vector<bool>(m, 0));

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

        for(int i=0;i<n;i++)
        {
            visited[i][0] = visited[i][m-1] = 1;
            pq.push({heights[i][0], i, 0});
            pq.push({heights[i][m-1], i, m-1});
        }

        for(int i=0;i<m;i++)
        {
            visited[0][i] = visited[n-1][i] = 1;
            pq.push({heights[0][i], 0, i});
            pq.push({heights[n-1][i], n-1, i});
        }

        int ans = 0;

        while(pq.size())
        {
            int h = pq.top()[0];
            int x = pq.top()[1];
            int y = pq.top()[2];
            pq.pop();

            for(int i=0;i<4;i++)
            {
                int r = x+dir[i][0];
                int c = y+dir[i][1];

                if(r>=0 && r<n && c>=0 && c<m && visited[r][c]==0)
                {
                    int maxH = max(heights[r][c], h);
                    ans += max(0, maxH-heights[r][c]);
                    visited[r][c] = 1;
                    pq.push({maxH, r, c});
                }
            }
        }
        return ans;
    }
    vector<vector<int>> dir = {{-1, 0}, {0, -1}, {0, 1}, {1, 0}};
};
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