<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
{
private:
    vector<int> size, parent, rank;

public:
    DisjointSet(int size)
    {
        this->size.resize(size + 1, 1);
        this->rank.resize(size + 1, 0);

        for (int i = 0; i <= size; i++)
        {
            this->parent.push_back(i);
        }
    }
    int findParent(int child)
    {
        if (parent[child] == child)
            return child;
        return parent[child] = findParent(parent[child]);
    }
    void unionBySize(int x, int y)
    {
        int xpar = findParent(x);
        int ypar = findParent(y);
        if (xpar == ypar)
            return;
        if (size[xpar] < size[ypar])
        {
            size[ypar] += size[xpar];
            parent[xpar] = ypar;
        }
        else
        {
            size[xpar] += size[ypar];
            parent[ypar] = xpar;
        }
    }
    void unionByRank(int x, int y)
    {
        int xpar = findParent(x);
        int ypar = findParent(y);
        if (xpar == ypar)
            return;
        if (rank[xpar] < rank[ypar])
        {
            parent[xpar] = ypar;
        }
        else if (rank[xpar] > rank[ypar])
        {
            parent[ypar] = xpar;
        }
        else
        {
            parent[ypar] = xpar;
            rank[xpar]++;
        }
    }
};


class Solution {
public:
    int makeConnected(int n, vector<vector<int>>& connections) {
        int extra=0;

        DisjointSet ds(n);
        for(auto con:connections){
            int u=con[0];
            int v=con[1];
            if(ds.findParent(u)!=ds.findParent(v)){
                ds.unionBySize(u,v);
            }
            else extra++;
           
        }
        
        int provinces=0;
        for(int i=0;i<n;i++){
            if(ds.findParent(i)==i)provinces++;
        }
// cout<<extra<<provinces;
        if(extra>=provinces-1)return provinces-1;
        return -1;

    }
};
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