<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:

    vector<vector<int>> directions = {{-1, 0}, {1, 0}, {0, 1}, {0, -1}};
    vector<vector<bool>> visited;

    int swimInWater(vector<vector<int>>& grid) {
        int n = grid.size();
        int low = 0, high = (n * n) - 1;
        while(low <= high) {
            visited = vector<vector<bool>> (n, vector<bool> (n, false));
            int mid = (high + low) >> 1;
            if(check(0, 0, n, grid, mid)) {
                high = mid - 1;
            } else {
                low = mid + 1;
            }
        }
        return low;
    }

    bool check(int x, int y, int n, vector<vector<int>>& grid, int time) {
        if(x < 0 || x >= n || y >= n || y < 0 || visited[x][y] || grid[x][y] > time)
            return 0;
        if(x == n - 1 && y == n - 1)
            return 1;
        visited[x][y] = 1;
        for(auto direction: directions) {
            int nx = x + direction[0], ny = y + direction[1];
            if(check(nx, ny, n, grid, time))
                return true;
        }
        return false;
    }
};
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