<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 m,n;
    vector<vector<int>>dir={{0,1},{0,-1},{1,0},{-1,0}};
    int swimInWater(vector<vector<int>>& arr) {
        m=arr.size();
        n=arr[0].size();
        int start=0;
        int end=m*n;
        int mid;
        int ans=INT_MAX;
        while(start<=end)
        {
            mid=(start+end)/2;
            if(fun(arr,mid))
            {
                ans=mid;
                end=mid-1;
            }
            else
            {
                start=mid+1;
            }
        }
        return ans;
    }
    bool fun(vector<vector<int>>&arr,int &path)
    {
        queue<vector<int>>q;
        q.push({0,0});
        vector<vector<bool>>vis(m,vector<bool>(n,false));
        vis[0][0]=true;
        if(arr[0][0]>path)
            return false;
        while(!q.empty())
        {
            int row=q.front()[0];
            int col=q.front()[1];
            q.pop();
            if(row==m-1 and col==n-1)
            {
                if(arr[row][col]>path)
                    return false;
                return true;
            }
            for(int i=0;i<dir.size();i++)
            {
                int r=row+dir[i][0];
                int c=col+dir[i][1];
                if((r>=0 and r<m) and (c>=0 and c<n))
                {
                    if(arr[r][c]<=path and vis[r][c]==false)
                    {
                        vis[r][c]=true;
                        q.push({r,c});
                    }
                }
            }
        }
        return false;
    }
};
23376