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Author: NinePiece2

Solutions/0794-swim-in-rising-water/README.md

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+<p>You are given an <code>n x n</code> integer matrix <code>grid</code> where each value <code>grid[i][j]</code> represents the elevation at that point <code>(i, j)</code>.</p>
+
+<p>It starts raining, and water gradually rises over time. At time <code>t</code>, the water level is <code>t</code>, meaning <strong>any</strong> cell with elevation less than equal to <code>t</code> is submerged or reachable.</p>
+
+<p>You can swim from a square to another 4-directionally adjacent square if and only if the elevation of both squares individually are at most <code>t</code>. You can swim infinite distances in zero time. Of course, you must stay within the boundaries of the grid during your swim.</p>
+
+<p>Return <em>the minimum time until you can reach the bottom right square </em><code>(n - 1, n - 1)</code><em> if you start at the top left square </em><code>(0, 0)</code>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2021/06/29/swim1-grid.jpg" style="width: 164px; height: 165px;" />
+<pre>
+<strong>Input:</strong> grid = [[0,2],[1,3]]
+<strong>Output:</strong> 3
+Explanation:
+At time 0, you are in grid location (0, 0).
+You cannot go anywhere else because 4-directionally adjacent neighbors have a higher elevation than t = 0.
+You cannot reach point (1, 1) until time 3.
+When the depth of water is 3, we can swim anywhere inside the grid.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2021/06/29/swim2-grid-1.jpg" style="width: 404px; height: 405px;" />
+<pre>
+<strong>Input:</strong> grid = [[0,1,2,3,4],[24,23,22,21,5],[12,13,14,15,16],[11,17,18,19,20],[10,9,8,7,6]]
+<strong>Output:</strong> 16
+<strong>Explanation:</strong> The final route is shown.
+We need to wait until time 16 so that (0, 0) and (4, 4) are connected.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>n == grid.length</code></li>
+	<li><code>n == grid[i].length</code></li>
+	<li><code>1 &lt;= n &lt;= 50</code></li>
+	<li><code>0 &lt;= grid[i][j] &lt;&nbsp;n<sup>2</sup></code></li>
+	<li>Each value <code>grid[i][j]</code> is <strong>unique</strong>.</li>
+</ul>

Solutions/0794-swim-in-rising-water/solution.py

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+class Solution:
+    def swimInWater(self, grid: List[List[int]]) -> int:
+        n = len(grid)
+        m = n ** 2
+        p = list(range(m))
+        high = [0] * m
+
+        def find(x: int) -> int:
+            if p[x] != x:
+                p[x] = find(p[x])
+            return p[x]
+
+        for i, row in enumerate(grid):
+            for j, h in enumerate(row):
+                high[h] = i * n + j
+
+        directions = (-1, 0, 1, 0, -1)
+        for t in range(m):
+            div, remain = divmod(high[t], n)
+            for dx, dy in pairwise(directions):
+                nx, ny = div + dx, remain + dy
+                if 0 <= nx < n and 0 <= ny < n and grid[nx][ny] <= t:
+                    p[find(div * n + remain)] = find(nx * n + ny)
+            
+            if find(0) == find(m - 1):
+                return t
+        
+        return 0