<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> size, parent, rank;

    DisjointSet(int n)
    {
        size.resize(n, 1);
        rank.resize(n, 0);
        parent.resize(n);
        for(int i = 0; i < n; i++) parent[i] = i;
    }

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

    void unionBySize(int u, int v)
    {
        int uPar = findParent(u), vPar = findParent(v);

        if(uPar == vPar) return;
        
        if(size[uPar] < size[vPar])
        {
            parent[uPar] = vPar;
            size[vPar] += size[uPar];
        }
        else
        {
            
            parent[vPar] = uPar;
            size[uPar] += size[vPar];
        }
    }

    void unionByRank(int u, int v)
    {
        int uPar = findParent(u), vPar = findParent(v);

        if(uPar == vPar) return;

        if(rank[uPar] < rank[vPar])
        {
            parent[uPar] = vPar;
        }
        else if(rank[vPar] < rank[uPar])
        {
            parent[vPar] = uPar; 
        }
        else
        {
            rank[uPar]++;
            parent[vPar] = uPar;
        }
    }
};

class Solution {
public:

int makeConnected(int n, vector<vector<int>>& connections)
{
    ios_base::sync_with_stdio(false);

    DisjointSet connect(n);
    int numCon = connections.size();
    int repeatedConnections = 0;
    for(auto it: connections)
    {
        int u = it[0], v = it[1];
        if(connect.findParent(u) == connect.findParent(v))
        {
            repeatedConnections++;
        }
        else
        {
            connect.unionBySize(u, v);
        }
    }

    int components = 0;
    for(int i = 0; i < n; i++)
    {
        if(connect.findParent(i) == i) components++;
    }
    
    if(repeatedConnections >= components-1) return components-1;
    return -1;
}
};
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