<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>The rain starts to fall. At time <code>t</code>, the depth of the water everywhere is <code>t</code>. 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 least 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>

class Solution {
public:
    int swimInWater(vector<vector<int>>& grid) {
        const int n = grid.size();

        vector<int> parents(n * n);
        iota(begin(parents), end(parents), 0);

        vector<tuple<int, int, int>> g;

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                int id = i * n + j;

                if (i > 0) {
                    g.emplace_back(id - n, id, max(grid[i][j], grid[i - 1][j]));
                }

                if (j > 0) {
                    g.emplace_back(id - 1, id, max(grid[i][j], grid[i][j - 1]));
                }
            }
        }

        sort(begin(g), end(g), [](auto& a, auto& b) {
            auto&& [a1, a2, a3] = a;
            auto&& [b1, b2, b3] = b;

            return a3 < b3;
        });

        function<int(int)> find = [&](int x) {
            while (x != parents[x]) {
                x = parents[x] = find(parents[x]);
            }

            return x;
        };

        auto connect = [&](int p, int q) {
            auto rootp = find(p);
            auto rootq = find(q);

            if (rootp == rootq) {
                return;
            }

            parents[rootp] = rootq;
        };

        for (auto& [a, b, c]: g) {
            connect(a, b);

            if (find(0) == find(n * n - 1)) {
                return c;
            }
        }

        return 0;
    }
};
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