<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:
  std::vector<std::pair<int, int>> dirs = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};

  bool check(const std::vector<std::vector<int>>& grid, int v) {
    std::set<std::pair<int, int>> used;
    std::queue<std::pair<int, int>> q;
    if (grid[0][0] > v) {
      return false;
    }
    q.push({0, 0});
    used.insert({0, 0});
    while (!q.empty()) {
      auto front = q.front();
      q.pop();
      for (auto it : dirs) {
        int x = front.first + it.first;
        int y = front.second + it.second;
        if (x >= 0 && x < grid.size() && y >= 0 && y < grid.size() && grid[x][y] <= v && used.find({x, y}) == used.end()) {
          q.push({x, y});
          used.insert({x, y});
        }
      }
    }
    return used.find({grid.size() - 1, grid.size() - 1}) != used.end();
  }

  int swimInWater(vector<vector<int>>& grid) {
    int l = -1;
    int r = 2500;
    while (l + 1< r) {
      int m = l + (r - l) / 2;
      if (check(grid, m)) {
        r = m;
      } else {
        l = m;
      }
    } 
    return r;
  }
};
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