<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 Solution {
public:
vector<int>parent;
vector<int>rank;
   int find(int x){
    if(x==parent[x]){
        return x;
    }
    return parent[x]=find(parent[x]);
   }

   void Union(int x,int y){
    int px=find(x);
    int py=find(y);
    if(px==py){
    return ;
    }else{
        if(rank[px]>rank[py]){
            parent[py]=px;
        }else if(rank[py]>rank[px]){
            parent[px]=py;
        }else{
            parent[px]=py;
            rank[py]++;
        }
    }
   }

void dfs(int source,int par, vector<vector<int>>&adj,vector<int>&vis){
    vis[source]=1;

    for(auto it:adj[source]){
        if(it!=par){
            if(!vis[it]){
dfs(it,source,adj,vis);
            }
            
    
        }
    }
}
    int makeConnected(int n, vector<vector<int>>& c) {
      
          parent.resize(n);
          rank.resize(n);
          for(int i=0;i<n;i++){
            rank[i]=0;
            parent[i]=i;
          }

        vector<vector<int>>adj(n);
        for (const auto& edge : c) {
            adj[edge[0]].push_back(edge[1]);
            adj[edge[1]].push_back(edge[0]);
        }

        vector<int>vis(n,0);
        int count=0;
        for(int i=0;i<n;i++){
            if(!vis[i]){
                dfs(i,-1,adj,vis);
                count++;
            }
        }
    //      int extras=0;
    // for (const auto& edge : c) {
    //        int a=edge[0];
    //         int b=edge[1];
    //        int para=find(a);
    //        int parb=find(b);
    //        if(para==parb){
    //         extras++;
    //        }else{
    //         Union(a,b);
    //        }
    //     }

//  if(extras>=(count-1)){
//     return count-1;
//  }
if(c.size()>=n-1){
    return count-1;
}
 return -1;

    }
};
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