[LeetCode Sync] Runtime - 27 ms (83.25%), Memory - 19.2 MB (79.71%)

Author: NinePiece2

Solutions/0417-pacific-atlantic-water-flow/README.md

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+<p>There is an <code>m x n</code> rectangular island that borders both the <strong>Pacific Ocean</strong> and <strong>Atlantic Ocean</strong>. The <strong>Pacific Ocean</strong> touches the island&#39;s left and top edges, and the <strong>Atlantic Ocean</strong> touches the island&#39;s right and bottom edges.</p>
+
+<p>The island is partitioned into a grid of square cells. You are given an <code>m x n</code> integer matrix <code>heights</code> where <code>heights[r][c]</code> represents the <strong>height above sea level</strong> of the cell at coordinate <code>(r, c)</code>.</p>
+
+<p>The island receives a lot of rain, and the rain water can flow to neighboring cells directly north, south, east, and west if the neighboring cell&#39;s height is <strong>less than or equal to</strong> the current cell&#39;s height. Water can flow from any cell adjacent to an ocean into the ocean.</p>
+
+<p>Return <em>a <strong>2D list</strong> of grid coordinates </em><code>result</code><em> where </em><code>result[i] = [r<sub>i</sub>, c<sub>i</sub>]</code><em> denotes that rain water can flow from cell </em><code>(r<sub>i</sub>, c<sub>i</sub>)</code><em> to <strong>both</strong> the Pacific and Atlantic oceans</em>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2021/06/08/waterflow-grid.jpg" style="width: 400px; height: 400px;" />
+<pre>
+<strong>Input:</strong> heights = [[1,2,2,3,5],[3,2,3,4,4],[2,4,5,3,1],[6,7,1,4,5],[5,1,1,2,4]]
+<strong>Output:</strong> [[0,4],[1,3],[1,4],[2,2],[3,0],[3,1],[4,0]]
+<strong>Explanation:</strong> The following cells can flow to the Pacific and Atlantic oceans, as shown below:
+[0,4]: [0,4] -&gt; Pacific Ocean 
+&nbsp;      [0,4] -&gt; Atlantic Ocean
+[1,3]: [1,3] -&gt; [0,3] -&gt; Pacific Ocean 
+&nbsp;      [1,3] -&gt; [1,4] -&gt; Atlantic Ocean
+[1,4]: [1,4] -&gt; [1,3] -&gt; [0,3] -&gt; Pacific Ocean 
+&nbsp;      [1,4] -&gt; Atlantic Ocean
+[2,2]: [2,2] -&gt; [1,2] -&gt; [0,2] -&gt; Pacific Ocean 
+&nbsp;      [2,2] -&gt; [2,3] -&gt; [2,4] -&gt; Atlantic Ocean
+[3,0]: [3,0] -&gt; Pacific Ocean 
+&nbsp;      [3,0] -&gt; [4,0] -&gt; Atlantic Ocean
+[3,1]: [3,1] -&gt; [3,0] -&gt; Pacific Ocean 
+&nbsp;      [3,1] -&gt; [4,1] -&gt; Atlantic Ocean
+[4,0]: [4,0] -&gt; Pacific Ocean 
+       [4,0] -&gt; Atlantic Ocean
+Note that there are other possible paths for these cells to flow to the Pacific and Atlantic oceans.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> heights = [[1]]
+<strong>Output:</strong> [[0,0]]
+<strong>Explanation:</strong> The water can flow from the only cell to the Pacific and Atlantic oceans.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>m == heights.length</code></li>
+	<li><code>n == heights[r].length</code></li>
+	<li><code>1 &lt;= m, n &lt;= 200</code></li>
+	<li><code>0 &lt;= heights[r][c] &lt;= 10<sup>5</sup></code></li>
+</ul>

Solutions/0417-pacific-atlantic-water-flow/solution.py

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+class Solution:
+    def pacificAtlantic(self, heights: List[List[int]]) -> List[List[int]]:
+        m, n = len(heights), len(heights[0])
+        visited1 = [[False] * n for _ in range(m)]
+        visited2 = [[False] * n for _ in range(m)]
+        queue1: Deque[Tuple[int, int]] = deque()
+        queue2: Deque[Tuple[int, int]] = deque()
+        directions = (-1, 0, 1, 0, -1)
+
+        for i in range(m):
+            queue1.append((i, 0))
+            visited1[i][0] = True
+            queue2.append((i, n - 1))
+            visited2[i][n - 1] = True
+        
+        for j in range(n):
+            queue1.append((0, j))
+            visited1[0][j] = True
+            queue2.append((m - 1, j))
+            visited2[m - 1][j] = True
+
+        def bfs(queue: Deque[Tuple[int, int]], visited: List[List[bool]]) -> None:
+            while queue:
+                x, y = queue.popleft()
+                for dx, dy in pairwise(directions):
+                    nx, ny = x + dx, y + dy
+
+                    if 0 <= nx < m and 0 <= ny < n and not visited[nx][ny] and heights[nx][ny] >= heights[x][y]:
+                        visited[nx][ny] = True
+                        queue.append((nx, ny))
+
+
+        bfs(queue1, visited1)
+        bfs(queue2, visited2)
+
+        return [(i, j) for i in range(m) for j in range(n) if visited1[i][j] and visited2[i][j]]