<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 f(vector<vector<int>>&grid,int i,int j,vector<vector<int>>&temp,int limit){
        int n=grid.size();
        queue<pair<int,int>>q;
        vector<int>a={1,0,-1,0};
        vector<int>b={0,-1,0,1};
        q.push({i,j});
        temp[i][j]=1;
        while(!q.empty()){
            int row=q.front().first;
            int col=q.front().second;
            if(row==n-1&&col==n-1)return true;
            q.pop();
            for(int i=0;i<4;i++){
                int nrow=row+a[i];
                int ncol=col+b[i];
                if(nrow>=0&&nrow<n&&ncol>=0&&ncol<n&&temp[nrow][ncol]!=1){
                    if(grid[nrow][ncol]>limit){
                       continue;
                    }
                    else{
                        temp[nrow][ncol]=1;
                        q.push({nrow,ncol});
                    }
                }
            }
        }
        return false;

    }
    int swimInWater(vector<vector<int>>& grid) {
        int n=grid.size();
        int low=0;
        int high=1e9;
        while(low<=high){
            vector<vector<int>>temp(n,vector<int>(n,-1));
            int mid=low+(high-low)/2;
            if(grid[0][0]<=mid&&f(grid,0,0,temp,mid)){
                high=mid-1;
            }
            else{
                low=mid+1;
            }
        }
        return low;
        
    }
};
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