<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>

// **ank**//
#include <bits/stdc++.h>
using namespace std;
#define ll                     long long int
#define all(v)                 v.begin(), v.end()
#define maxi(v)                *max_element(v.begin(), v.end())
#define vecsum(v)              accumulate(v.begin(), v.end(), 0LL)
#define mini(v)                *min_element(v.begin(), v.end())
#define sortd(v)                sort(v.begin(), v.end(), greater<int>())
#define read(a, n)             for(int i=0;i<n;i++)    cin >> a[i]
#define print(a, n)            for(int i=0;i<n;i++)    cout << a[i] << " "; cout << endl
#define rep(a,n,b,i)           for(int i=a;i<n;i+=b)       
#define M                       1000000007
#define pb                      push_back
#define sz                      size()
#define len                     length()
#define fir                     first
#define sec                     second
#define cy                      cout << "YES" << endl
#define cn                      cout << "NO" << endl 
#define mll                     map<ll,ll>
#define mcl                     map<char,ll>
#define msl                     map<string,ll>
#define si                      set<ll>
#define sc                      set<char>
using pp = pair<ll,ll>;
using vl = vector<ll>;
using vc = vector<char>;
using vi = vector<int>;
using vvl=vector<vector<ll>>;
using vvc=vector<vector<char>>;
class Solution {
public:
void dfs(ll node,vvl &adj,vl&vis){
    vis[node]=1;
    for(auto it:adj[node])if(!vis[it])dfs(it,adj,vis);
    return ;
}
    int makeConnected(int n, vector<vector<int>>& con) {
        vector<ll>vis(n,0);
        vvl adj(n);
        for(auto it:con){
           adj[it[0]].pb(it[1]);
           adj[it[1]].pb(it[0]);
        }
        ll ct=0,ed=con.sz;
         ll rem=ed-n+1;
        for(int i=0;i<n;i++)
        if(!vis[i]){
            ct++;
            dfs(i,adj,vis);
        }
        if(ed<n-1)return -1;
        return ct-1;
    }
};
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