<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:
    bool IsReachable(const std::vector<std::vector<int>>& grid, int n, int t) {
        if (grid[0][0] > t) {
            return false;
        }
        
        const std::vector<int> dy {-1, 1, 0, 0};
        const std::vector<int> dx {0, 0, -1, 1};
        std::vector<std::vector<bool>> visited(n, std::vector<bool>(n, false));
        std::queue<std::pair<int, int>> q;
        q.push({0, 0});
        while (!q.empty()) {
            const auto cur = q.front();
            q.pop();

            const int y = cur.first;
            const int x = cur.second;

            if (visited[y][x]) {
                continue;
            }
            visited[y][x] = true;

            if (y == n-1 && x == n-1) {
                break;
            }

            for (int i = 0; i < dy.size(); ++i) {
                const int yy = y + dy[i];
                const int xx = x + dx[i];

                if (yy >= 0 && yy < n && xx >= 0 && xx < n && grid[yy][xx] <= t) {
                    q.push({yy, xx});
                }
            }
        }

        return visited[n-1][n-1];
    }

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

        int mx = grid[0][0];
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                mx = std::max(mx, grid[i][j]);
            }
        }

        int l = -1, r = mx;
        while (r - l > 1) {
            int mid = (r + l) / 2;
            if (IsReachable(grid, n, mid)) {
                r = mid;
            } else {
                l = mid;
            }
        }

        return r;
    }
};
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