<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 find(vector<int> &parent,int n){
        if(parent[n]==n)return n;
        return parent[n]=find(parent,parent[n]);
    }
    void unionbyrank(vector<int> &parent,vector<int>&rank,int u,int v){
        int ulp_u=find(parent,u);
        int ulp_v=find(parent,v);
        if(rank[ulp_u]<rank[ulp_v])parent[ulp_u]=ulp_v;
        else if(rank[ulp_u]>rank[ulp_v])parent[ulp_v]=ulp_u;
        else{
            rank[ulp_u]++;
            parent[ulp_v]=ulp_u;
        }
    }
    int swimInWater(vector<vector<int>>& grid) {
        int n=grid.size();
        int m=n*n;
        vector<int> parent(m+1,0),rank(m+1,0);
        vector<pair<int,pair<int,int>>> edges;
        vector<vector<int>> dir={{0,1},{0,-1},{1,0},{-1,0}};
        for(int i=0;i<m;i++)parent[i]=i;
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                int node=i*n+j;
                for(auto it:dir){
                    int i1=i+it[0];
                    int j1=j+it[1];
                    if(i1>=0&&i1<n&&j1>=0&&j1<n)
                    {int adjnode=i1*n+j1;
                    edges.push_back({max(grid[i][j],grid[i1][j1]),{node,adjnode}});}
                }
            }
        }
        int mxwt=0;
        sort(edges.begin(),edges.end());
        for(auto it:edges){
            if(find(parent,(n-1)*n+(n-1))==find(parent,0))return mxwt;
            int wt=it.first;
            int node=it.second.first;
            int adjnode=it.second.second;
            if(find(parent,node)!=find(parent,adjnode)){
                mxwt=max(mxwt,wt);
                unionbyrank(parent,rank,node,adjnode);
            }
        }
        return mxwt;
    }
};
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