<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) {
        // water can only be trapped inside the height map, not at the boundary
        // so all the boundaries are marked as visited
        if (heightMap.empty() || heightMap[0].empty()) {
            return 0;
        }
        int row = heightMap.size();
        int col = heightMap[0].size();

        vector<vector<bool>> visited(row, vector<bool>(col, false));

        // pq stores the height and location of each cell
        priority_queue<pair<int, pair<int, int>>, vector<pair<int, pair<int, int>>>, greater<pair<int, pair<int, int>>>> pq;

        // push all boundary cells into the pq
        for (int i = 0; i < row; ++i) {
            pq.push({heightMap[i][0], {i, 0}});
            pq.push({heightMap[i][col - 1], {i, col - 1}});
            visited[i][0] = true;
            visited[i][col - 1] = true;
        }
        for (int j = 1; j < col - 1; ++j) {
            pq.push({heightMap[0][j], {0, j}});
            pq.push({heightMap[row - 1][j], {row - 1, j}});
            visited[0][j] = true;
            visited[row - 1][j] = true;
        }

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

        int waterTrapped = 0;

        while (!pq.empty()) {
            auto [height, cell] = pq.top();
            pq.pop();
            int x = cell.first;
            int y = cell.second;

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

                if (nx < 0 || nx >= row || ny < 0 || ny >= col || visited[nx][ny]) {
                    continue;
                }

                // if the neighbour's height is smaller, water can be trapped
                if (heightMap[nx][ny] < height) {
                    waterTrapped += height - heightMap[nx][ny];
                }

                pq.push({max(height, heightMap[nx][ny]), {nx, ny}});
                visited[nx][ny] = true;
            }
        }
        return waterTrapped;
    }
};
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