<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:
    
    int find_par(int u,vector<int>& par){
        if(u == par[u]) return u;
        else return par[u] = find_par(par[u],par);
    }
    void make_union(int u,int v,vector<int>& par,vector<int>& rank){
        int pu = find_par(u,par);
        int pv = find_par(v,par);
        if(rank[pu] > rank[pv]){
            par[pv] = pu;
        }else if(rank[pv] > rank[pu]){
            par[pu] = pv;
        }else {
            par[pu] = pv;
            rank[pv]++;
        }
        return ;
    }
    int makeConnected(int n, vector<vector<int>>& connections) {
     // Solution obj(n);
     vector<int> rank(n,0);
    vector<int> par(n,0);
   
     
        for(int i=0;i<n;i++) par[i]=i;
    
      int edges=0;
      int size = connections.size();
      set<int> ss;
      for(int i=0;i<size;i++){
        if(find_par(connections[i][0],par)!=find_par(connections[i][1],par)) {
            int u = connections[i][0];
            int v = connections[i][1];
            make_union(u,v,par,rank);
            ss.insert(u);
            ss.insert(v);
        }else{
            edges++;
        }
      }
      if(n-ss.size() > edges) return -1;
      int ans=0;
      for(int i=0;i<n;i++){
        if(i == par[i]) ans++;
      } 
      return ans-1;
    }
};
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