<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:
    struct unionfind
    {
        vector<int>par,rank;
        unionfind(int n)
        {
            par.resize(n);
            rank.resize(n);
            for(int i=0;i<n;i++)
            {
                par[i]=i;
                rank[i]=1;
            }
        }
        int find(int x)
        {
            if(par[x]==x)
                return x;
            return par[x]=find(par[x]);
        }
        void merge(int x, int y)
        {
            int px=find(x);
            int py=find(y);
            if(px!=py)
            {
                if(rank[px]>rank[py])
                    swap(px,py);
                par[px]=py;
                rank[py]+=rank[px];
            }
        }
    };
    bool check(int mid, vector<vector<int>>& grid)
    {
        int n=grid.size();
        unionfind uf(n*n);
        queue<pair<int,int>>q;
        int dx[]={0,0,1,-1};
        int dy[]={1,-1,0,0};
        if(grid[0][0]<=mid)
        q.push({0,0});
        else return false;
        vector<vector<int>>vis(n,vector<int>(n,0));
        vis[0][0]=1;
        while(!q.empty())
        {
            int x=q.front().first;
            int y=q.front().second;
            q.pop();
            for(int i=0;i<4;i++)
            {
                int nx=x+dx[i];
                int ny=y+dy[i];
                if(nx>=0 && nx<n && ny>=0 && ny<n && !vis[nx][ny] && grid[nx][ny]<=mid)
                {
                    vis[nx][ny]=1;
                    q.push({nx,ny});
                    uf.merge(x*n+y,nx*n+ny);
                }
            }
        }
        return uf.find(0)==uf.find(n*n-1);
    }
    int swimInWater(vector<vector<int>>& grid) {
        int n=grid.size();
        int l=0;
        int r=n*n-1;
        int ans=-1;
        while(l<=r)
        {
            int mid=l+(r-l)/2;
            if(check(mid,grid))
            {
                ans=mid;
                r=mid-1;
            }
            else
            {
                l=mid+1;
            }
        }
        return ans;
    }
};
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