<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:
    const vector<vector<int>> dirs = {{0, 1}, {1, 0}, {-1, 0}, {0, -1}};
    int swimInWater(vector<vector<int>>& grid) {
        int n = grid.size();
        auto check = [&](int t) {
            if(grid[0][0] > t) {
                return false;
            }
            queue<vector<int>> q;
            vector<vector<bool>> vis(n, vector<bool>(n, false));
            q.push({0, 0});
            vis[0][0] = true;
            while(!q.empty()) {
                int qs = q.size();
                while(qs--) {
                    int x = q.front()[0], y = q.front()[1];
                    q.pop();
                    for(auto &dir: dirs) {
                        int nx = x + dir[0], ny = y + dir[1];
                        if(nx >= 0 && nx < n && ny >= 0 && ny < n && !vis[nx][ny] && grid[nx][ny] <= t) {
                            vis[nx][ny] = true;
                            q.push({nx, ny});
                        }
                    }
                }
            }
            return vis[n - 1][n - 1] == true;
        };
        int l = 0, r = n * n + 5, ans = -1;
        while(l <= r) {
            int mid = (l + r) / 2;
            if(check(mid)) {
                ans = mid;
                r = mid - 1;
            }
            else {
                l = mid + 1;
            }
        }
        return ans;
    }
};
23175