[LeetCode Sync] Runtime - 132 ms (51.98%), Memory - 20.5 MB (37.70%)

Author: NinePiece2

Solutions/0407-trapping-rain-water-ii/README.md

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+<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>

Solutions/0407-trapping-rain-water-ii/solution.py

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+class Solution:
+    def trapRainWater(self, heightMap: List[List[int]]) -> int:
+        m, n = len(heightMap), len(heightMap[0])
+        visited = [[False] * n for _ in range(m)]
+        priority_queue = []
+
+        for i in range(m):
+            for j in range(n):
+                if i == 0 or i == m - 1 or j == 0 or j == n - 1:
+                    heappush(priority_queue, (heightMap[i][j], i, j))
+                    visited[i][j] = True
+        
+        directions = (-1, 0, 1, 0, -1)
+        result = 0
+
+        while priority_queue:
+            height, i, j = heappop(priority_queue)
+            for dir1, dir2 in pairwise(directions):
+                x, y = i + dir1, j + dir2
+                if x >= 0 and x < m and y >= 0 and y < n and not visited[x][y]:
+                    result += max(0, height - heightMap[x][y])
+                    heappush(priority_queue, (max(height, heightMap[x][y]), x, y))
+                    visited[x][y] = True
+
+        return result