<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:
    
    int n;
    vector<int> parent;
    vector<int> size;

    void make(int n)
    {
        
        for(int i=0; i<n; i++)
        {
            parent.push_back(i);
            size.push_back(1);
        }
    }

    int find(int n)
    {
        if(parent[n]==n)
        return n;
        
        return parent[n] = find(parent[n]);
    }

    void Union(int a, int b)
    {
        int pa = find(a);
        int pb = find(b);

        if(pa != pb)
        {
            if(size[pa] < size[pb])
            swap (pa,pb);

            size[pa]+= size[pb];
            parent[pb] = pa; 

        }
    }

};

class Solution {
public:
    int makeConnected(int n, vector<vector<int>>& connections) {
        int nodes = n;
        int edges = connections.size();

        if(edges < nodes-1)
        return -1;

        int cnt = 0;
        int extra = 0;

        DSU s1;
        s1.make(nodes);
        
        for(auto &it: connections)
        {
            int v1 = it[0];
            int v2 = it[1];

            if(s1.find(v1)==s1.find(v2))
            {
                extra++;
            }
            else
            {
                s1.Union(v1, v2);
            }
        }

        for(int i=0; i<nodes; i++)
        {
            if(s1.find(i) == i)
            cnt++;
        }

        if(cnt-1 <= extra)
        return cnt-1;

        return -1;
        


    }
};
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