<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 bfs(vector<vector<int>>& grid , int mid){
        set<pair<int,int>> st;
        queue<pair<int,int>> q;
        int n = grid.size();

        q.push({0,0});
        st.insert({0,0});
        while(!q.empty()){
            auto[x , y] = q.front();
            q.pop();
            if(x==n-1 && y==n-1) return 1;
            // if(st.count({x,y})>0) continue;
            int dx[] = {0,0,-1,1};
            int dy[] = {1,-1,0,0};
            for(int i = 0 ; i < 4 ; i++){
                int nx = x+dx[i];
                int ny = y+dy[i];
                if(ny>=0 && ny<n && nx>=0 && nx<n && grid[nx][ny] <= mid){
                    if(st.count({nx,ny})==0){
                        q.push({nx,ny});
                        st.insert({nx,ny});
                    }
                }
            }
        }

        return 0;
        
    }
    int swimInWater(vector<vector<int>>& grid) {
        int n = grid.size();
        int l = grid[n-1][n-1];
        int r = 0;
        for(int i = 0 ; i < n ; i++){
            int mx = *max_element(grid[i].begin() , grid[i].end());
            r = max(r , mx);
        }
        if(r<=l) return l;

        while(l<=r){
            int mid =  l + (r-l)/2;
            if(grid[0][0] > mid ) {
                l = grid[0][0];
            }
            if(bfs(grid , mid)){
                r = mid-1;
            }else{
                l = mid+1;
            }
        }
        return l;
    }
};
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