Swim in Rising Water
<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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0700-0799/0778.Swim%20in%20Rising%20Water/images/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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0700-0799/0778.Swim%20in%20Rising%20Water/images/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>
Depth-First Search; Breadth-First Search; Union Find; Array; Binary Search; Matrix; Heap (Priority Queue)
class Solution:
    def swimInWater(self, grid: List[List[int]]) -> int:
        def find(x):
            if p[x] != x:
                p[x] = find(p[x])
            return p[x]

        n = len(grid)
        p = list(range(n * n))
        hi = [0] * (n * n)
        for i, row in enumerate(grid):
            for j, h in enumerate(row):
                hi[h] = i * n + j
        for t in range(n * n):
            i, j = hi[t] // n, hi[t] % n
            for a, b in [(0, -1), (0, 1), (1, 0), (-1, 0)]:
                x, y = i + a, j + b
                if 0 <= x < n and 0 <= y < n and grid[x][y] <= t:
                    p[find(x * n + y)] = find(hi[t])
                if find(0) == find(n * n - 1):
                    return t
        return -1
