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 {
public:
    vector<int> p;

    int swimInWater(vector<vector<int>>& grid) {
        int n = grid.size();
        p.resize(n * n);
        for (int i = 0; i < p.size(); ++i) p[i] = i;
        vector<int> hi(n * n);
        for (int i = 0; i < n; ++i)
            for (int j = 0; j < n; ++j)
                hi[grid[i][j]] = i * n + j;
        vector<int> dirs = {-1, 0, 1, 0, -1};
        for (int t = 0; t < n * n; ++t) {
            int i = hi[t] / n, j = hi[t] % n;
            for (int k = 0; k < 4; ++k) {
                int x = i + dirs[k], y = j + dirs[k + 1];
                if (x >= 0 && x < n && y >= 0 && y < n && grid[x][y] <= t)
                    p[find(x * n + y)] = find(hi[t]);
                if (find(0) == find(n * n - 1)) return t;
            }
        }
        return -1;
    }

    int find(int x) {
        if (p[x] != x) p[x] = find(p[x]);
        return p[x];
    }
};