<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> parent;
    vector<int> rank;
    
    DisjointSet(int n){
        parent.resize(n);
        rank.resize(n,0);
        for(int i=0;i<n;i++){
            parent[i] = i;// self parent
        }
    }

    int findParent(int node){
        if(parent[node]==node) return node;
        return parent[node] = findParent(parent[node]);
    }

    void unionByRank(int u, int v){
        int pu = findParent(u);
        int pv = findParent(v);
        
        if(pu==pv) return ;
        
        int ru = rank[pu];
        int rv = rank[pv];
        
        if(ru<=rv){
            parent[pu] = pv;
            rank[pv] += (ru==rv)?1:0;
        }else{
            parent[pv] = pu;
        }
    }
};
class Solution {
public:
    int makeConnected(int n, vector<vector<int>>& conn) {
        if(conn.size()<n-1) return -1;

        DisjointSet ds(n);
        unordered_set<int> s;
        // int up=-1;
        for(auto c: conn){
            int src = c[0];
            int dest = c[1];
            
            if(ds.findParent(src)!=ds.findParent(dest)) {
                ds.unionByRank(src, dest);
            }
        }
        // int up = ds.findParent(conn[0][0]);
        // cout<<" selected"
        // int ans=0;
        for(int i=0;i<n;i++){
            // cout<<" parent of "<< i <<" : "<<ds.findParent(i)<<endl;
            // if(ds.findParent(i)!=up) ans++;
            s.insert(ds.findParent(i));
        }
        return s.size()-1;
    }
};
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