<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) {
        std::set<std::pair<int, int>> visited;
        std::priority_queue<std::vector<int>, std::vector<std::vector<int>>,
                            std::greater<>>
            queue;
        std::vector<std::vector<int>> dir = {{0, 1}, {1, 0}, {-1, 0}, {0, -1}};
        queue.push({grid[0][0], 0, 0});
        visited.insert({0, 0});
        while (!queue.empty()) {
            auto cur = queue.top();
            queue.pop();
            if (cur[1] == grid.size() - 1 && cur[2] == grid.size() - 1)
                return cur[0];
            for (auto d : dir) {
                if (cur[1] + d[0] < 0 || cur[2] + d[1] < 0 ||
                    cur[1] + d[0] >= grid.size() ||
                    cur[2] + d[1] >= grid.size() ||
                    visited.count({cur[1] + d[0], cur[2] + d[1]}))
                    continue;
                visited.insert({cur[1] + d[0], cur[2] + d[1]});
                queue.push(
                    {std::max(cur[0], grid[cur[1] + d[0]][cur[2] + d[1]]),
                     cur[1] + d[0], cur[2] + d[1]});
            }
        }
        return -1;
    }
};

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