<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 DSU{
    public:
    vector<int> rank,par,size;
    int len=0;
    DSU(int n){
        rank.resize(n,1);
        size.resize(n,1);
        par.resize(n);
        for(int i=0;i<n;i++){
            par[i]=i;
        }
        len=n;
    }
    
    int find(int x){
        if(par[x]==x)   return x;
        return par[x] = find(par[x]);
    }
    
    void UnionBySize(int a,int b){
        a=find(a);
    	b=find(b);
    
        if(size[a]<=size[b]){
            size[b]+=size[a];
            par[a]=b;
        }
        else{
            size[a]+=size[b];
            par[b]=a;
        }
    }
    
    void UnionByRank(int a,int b){
        a=find(a);
    	b=find(b);
    
        if(rank[a]<=rank[b]){
            par[a]=b;
            rank[b]++;
        }
        else{
            par[b]=a;
            rank[a]++;
        }
    }
};

class Solution {
public:
    int makeConnected(int n, vector<vector<int>>& conn) {

        DSU dsu(n);
        int count=0;
        for(int i=0;i<conn.size();i++){
            int u=conn[i][0];
            int v=conn[i][1];
;                                      
            if(dsu.find(u)!=dsu.find(v)){
                dsu.UnionByRank(u,v);
            }
            else    count++;
        }
        vector<int> parent=dsu.par;
        int components=0;
        for(int i=0;i<n;i++){
            if(parent[i]==i)   components++;
        }

        int ans=components-1;

        // cout<<total<<" "<<count;
        if(ans>count)     return -1;
        return ans;
    }
};
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