<p>There are <code>n</code> computers numbered from <code>0</code> to <code>n - 1</code> connected by ethernet cables <code>connections</code> forming a network where <code>connections[i] = [a<sub>i</sub>, b<sub>i</sub>]</code> represents a connection between computers <code>a<sub>i</sub></code> and <code>b<sub>i</sub></code>. Any computer can reach any other computer directly or indirectly through the network.</p>

<p>You are given an initial computer network <code>connections</code>. You can extract certain cables between two directly connected computers, and place them between any pair of disconnected computers to make them directly connected.</p>

<p>Return <em>the minimum number of times you need to do this in order to make all the computers connected</em>. If it is not possible, return <code>-1</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/01/02/sample_1_1677.png" style="width: 500px; height: 148px;" />
<pre>
<strong>Input:</strong> n = 4, connections = [[0,1],[0,2],[1,2]]
<strong>Output:</strong> 1
<strong>Explanation:</strong> Remove cable between computer 1 and 2 and place between computers 1 and 3.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/01/02/sample_2_1677.png" style="width: 500px; height: 129px;" />
<pre>
<strong>Input:</strong> n = 6, connections = [[0,1],[0,2],[0,3],[1,2],[1,3]]
<strong>Output:</strong> 2
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> n = 6, connections = [[0,1],[0,2],[0,3],[1,2]]
<strong>Output:</strong> -1
<strong>Explanation:</strong> There are not enough cables.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= connections.length &lt;= min(n * (n - 1) / 2, 10<sup>5</sup>)</code></li>
	<li><code>connections[i].length == 2</code></li>
	<li><code>0 &lt;= a<sub>i</sub>, b<sub>i</sub> &lt; n</code></li>
	<li><code>a<sub>i</sub> != b<sub>i</sub></code></li>
	<li>There are no repeated connections.</li>
	<li>No two computers are connected by more than one cable.</li>
</ul>

class DisjointSet{
    
    public:
    vector<int> rank,parent,size;
    DisjointSet(int n){
        rank.resize(n+1,0);
        parent.resize(n+1,0);
        size.resize(n+1,0);
        for(int i=0;i<=n;i++){
            parent[i]=i;
            size[i]=1;
        }
    }
    int findUPar(int node){
        if(node==parent[node]) return node;
        return parent[node]=findUPar(parent[node]);
    }

    void unionByRank(int u,int v){
        int ulp_v=findUPar(v);
        int ulp_u=findUPar(u);

        if(ulp_u==ulp_v) return ;
        if(rank[ulp_u]<rank[ulp_v]){
            parent[ulp_u]=ulp_v;
        }
        else if(rank[ulp_v]<rank[ulp_u]){
            parent[ulp_v]=ulp_u;
        }

        else{
            parent[ulp_v]=ulp_u;
            rank[ulp_u]++;
        }
    }
     void unionBySize(int u,int v){
        int ulp_v=findUPar(v);
        int ulp_u=findUPar(u);
         if(ulp_u==ulp_v) return ;
         if(size[ulp_u]<size[ulp_v]){
            parent[ulp_u]=ulp_v;
            size[ulp_v]+=size[ulp_u];
         }
         else{
            parent[ulp_v]=ulp_u;
            size[ulp_u]+=size[ulp_v];
         }



     }

};
class Solution {
public:
    int makeConnected(int n, vector<vector<int>>& edges) {
        DisjointSet ds(n);
        int ext=0;
        for(auto it:edges){
            int u=it[0];
            int v=it[1];
            if(ds.findUPar(u)==ds.findUPar(v)){
                ext++;
            }
            else ds.unionBySize(u,v);
        }
      
        
          int ret=0;
        for(int i=0;i<n;i++){
            if(ds.parent[i]==i)ret++;
        }
        if(ext>=ret-1) return ret-1;
        return -1;
    }
};
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