<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:
    class Disjoint{
    public:
        vector<int> rank,size,parent;
        Disjoint(int n)
        {
            rank.resize(n,0);
            size.resize(n,1);
            parent.resize(n);
            for(int i=0;i<n;i++)
            parent[i]=i;
        }
        int find(int x){
            if(x==parent[x])
            return x;
            else return find(parent[x]);
        }

        void UnionR(int x, int y)
        {
            int xp=find(x),yp=find(y);
            if(xp==yp)
            return;
            if(rank[xp]>rank[yp])
            {
                parent[yp]=xp;
            }
            else if(rank[xp]<rank[yp])
            {
                parent[xp]=yp;
            }
            else{
                parent[yp]=xp;
                rank[xp]++;
            }
        }

        void UnionS(int x, int y)
        {
            int xp=find(x),yp=find(y);
            if(xp==yp)
            return;
            if(size[xp]>=size[yp])
            {
                parent[yp]=xp;
                size[xp]+=size[yp];
            }
            else{
                parent[xp]=yp;
                size[yp]+=size[xp];
            }
        }
    };

    int makeConnected(int n, vector<vector<int>>& connections) {
        Disjoint network(n);
        for(auto it: connections)
        {
            network.UnionS(it[0],it[1]);
        }
        for(int i=0;i<n;i++)
        network.parent[i]=network.find(i);
        unordered_map<int,int> mp,cnt;
        unordered_set<int> st;
        for(int i=0;i<n;i++)
        {
            mp[network.parent[i]]++;
            st.insert(network.parent[i]);
        }
        for(auto it: connections)
        {
            cnt[network.parent[it[0]]]++;
            cnt[network.parent[it[1]]]++;
        }
        int tot=0;
        for(auto it: st)
        {
            cnt[it]/=2;
            tot+=max(cnt[it]+1-mp[it],0);
        }
        if(tot+1>=st.size())
        return st.size()-1;
        else return -1; 
    }
};
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