<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 m = heightMap.size();
        if (m == 0) return 0;
        int n = heightMap[0].size();
        if (n == 0) return 0;
        vector<vector<bool>> visited(m, vector<bool>(n, false));
        priority_queue<pair<int, pair<int, int>>, vector<pair<int, pair<int, int>>>, greater<pair<int, pair<int, int>>>> pq;

        // Push all the boundary cells into the priority queue
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (i == 0 || j == 0 || i == m - 1 || j == n - 1) {
                    pq.push({heightMap[i][j], {i, j}});
                    visited[i][j] = true;
                }
            }
        }

        int result = 0;
        vector<int> dirs = {0, 1, 0, -1, 0};

        while (!pq.empty()) {
            auto curr = pq.top(); pq.pop();
            int height = curr.first;
            int x = curr.second.first;
            int y = curr.second.second;

            for (int k = 0; k < 4; ++k) {
                int nx = x + dirs[k];
                int ny = y + dirs[k + 1];

                if (nx >= 0 && nx < m && ny >= 0 && ny < n && !visited[nx][ny]) {
                    visited[nx][ny] = true;
                    result += max(0, height - heightMap[nx][ny]);
                    pq.push({max(height, heightMap[nx][ny]), {nx, ny}});
                }
            }
        }

        return result;
    }
};
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