<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>

#define pii pair<int,int>

struct pair_hash {
    inline std::size_t operator()(const std::pair<int,int> & v) const {
        return v.first*31+v.second;
    }
};

vector<int> delta = {0, 1, 0, -1, 0};
class Solution {
    bool check (vector<vector<int>>& A, int mx) {
        int n = A.size();
        queue<pii> q;
        unordered_set<pii, pair_hash> queued;
        if (A[0][0] <= mx) {
            q.push({0,0});
            queued.insert({0,0});
        }  

        while (q.size()) {
            int m=q.size();
            while (m--) {
                auto [i, j] = q.front();    q.pop();
                if (i==n-1 and j==n-1)  return true;
                for (int d=0 ; d<4 ; d++) {
                    int ii = i + delta[d], jj = j + delta[d+1];
                    if (
                        min(ii,jj) >= 0 and 
                        max(ii,jj) < n and
                        A[ii][jj] <= mx and 
                        not queued.count({ii,jj})

                    ) {
                        q.push({ii, jj});
                        queued.insert({ii,jj});
                    }

                }

            }
        }

        return false;
    }
public:
    int swimInWater(vector<vector<int>>& A) {
        int n=A.size();
        int l=0, r=n*n;
        while (l<r) {
            int m = l+(r-l)/2;
            if (check(A,m)) r = m;
            else l = m+1;
        }
        return l;
    }
};
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