<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>>& heightMap) {
        int n = heightMap.size();
        int m = heightMap[0].size();
        priority_queue<vector<int>, vector<vector<int>>, greater<vector<int>>> pq;
        vector<vector<int>> visited(n, vector<int>(m, 0));
        for (int i=0; i<n; i++) {
            pq.push({heightMap[i][0], i, 0});
            pq.push({heightMap[i][m-1], i, m-1});
            visited[i][0] = 1;
            visited[i][m-1] = 1;
        }  
        for (int j=0; j<m; j++) {
            pq.push({heightMap[0][j], 0, j});
            pq.push({heightMap[n-1][j], n-1, j});
            visited[0][j] = 1;
            visited[n-1][j] = 1;
        }
        int ans = 0;
        while(!pq.empty()) {
            int h = pq.top()[0];
            int i = pq.top()[1];
            int j = pq.top()[2];
            ans += max(0, h-heightMap[i][j]);
            pq.pop();

            vector<pair<int, int>> dir{{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
            for (pair<int, int> d : dir) {
                int ni = d.first + i;
                int nj = d.second + j;
                if (isValid(ni, nj, n, m) && visited[ni][nj] != 1) {
                    int nh = max(heightMap[ni][nj], h);
                    pq.push({nh, ni, nj});
                    visited[ni][nj] = 1;
                }
            }
        }
        return ans; 
    }

    bool isValid(int i, int j, int n, int m) {
        return (0 <= i && i < n && 0 <= j && j < m) ? true : false;
    }
};
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