<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:

    int canReach(vector<vector<int>>& grid, int t, int n){
        unordered_set<int> visited; 
        queue<int> q; 
        int offsets[4][2] = {{0, 1}, {0, -1}, {-1, 0}, {1, 0}};
        int i = 0, j = 0, next_i, next_j; 
        q.push(i * n + j);
        visited.insert(i * n + j);
        while(!q.empty()){
            auto front = q.front();
            q.pop();
            i = front / n;
            j = front % n;
            if(i == n - 1 && j == n - 1) return true; 
            for(auto offset: offsets){
                next_i = i + offset[0];
                next_j = j + offset[1];
                if(next_i >= 0 && next_i < n && next_j >=0 && next_j < n && grid[next_i][next_j] <= t &&
                visited.find(next_i * n + next_j) == visited.end()){
                    q.push({next_i * n + next_j});
                    visited.insert(next_i * n + next_j);
                }
            }
        }
        return false; 
    }


    int swimInWater(vector<vector<int>>& grid) {
        int start = grid[0][0], end = 0, n = grid.size(); 
        if(canReach(grid, start, n)) return start; 
        for(int i = 0; i < n ; i++){
            for(int j = 0; j < n; j++){
                end = max(end, grid[i][j]);
            }
        }
        while(end - start > 1){
            int mid = start + (end - start) / 2; 
            if(canReach(grid, mid, n)){
                end = mid;
            }else{
                start = mid;
            }
        }
        return end;
    }
};
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