Trapping Rain Water II
<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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0400-0499/0407.Trapping%20Rain%20Water%20II/images/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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0400-0499/0407.Trapping%20Rain%20Water%20II/images/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>
Breadth-First Search; Array; Matrix; Heap (Priority Queue)
class Solution {
    public int trapRainWater(int[][] heightMap) {
        int m = heightMap.length, n = heightMap[0].length;
        boolean[][] vis = new boolean[m][n];
        PriorityQueue<int[]> pq = new PriorityQueue<>((a, b) -> a[0] - b[0]);
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (i == 0 || i == m - 1 || j == 0 || j == n - 1) {
                    pq.offer(new int[] {heightMap[i][j], i, j});
                    vis[i][j] = true;
                }
            }
        }
        int ans = 0;
        int[] dirs = {-1, 0, 1, 0, -1};
        while (!pq.isEmpty()) {
            var p = pq.poll();
            for (int k = 0; k < 4; ++k) {
                int x = p[1] + dirs[k], y = p[2] + dirs[k + 1];
                if (x >= 0 && x < m && y >= 0 && y < n && !vis[x][y]) {
                    ans += Math.max(0, p[0] - heightMap[x][y]);
                    vis[x][y] = true;
                    pq.offer(new int[] {Math.max(p[0], heightMap[x][y]), x, y});
                }
            }
        }
        return ans;
    }
}