<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 Disjoint_set
{

public:
    vector<int> rank, parent;
    Disjoint_set(int n)
    {
        rank.resize(n+1 , 0);
        parent.resize(n+1 );
        for (int i = 0; i <=n; i++)
        {
            parent[i] = i;
        }
    }
    int find_uparent(int node)
    {
        if (parent[node] == node)
            return node;
        return parent[node] = find_uparent(parent[node]);
    }
    void unionByRank(int u, int v)
    {
        int parent_u = find_uparent(u);
        int parent_v = find_uparent(v);
        if (parent_u == parent_v)
        {
            return;
        }
        if (rank[parent_u] < rank[parent_v])
        {
            parent[parent_u] = parent_v;
        }
        else if (rank[parent_v] < rank[parent_u])
        {
            parent[parent_v] = parent_u;
        }
        else
        {
            parent[parent_v] = parent_u;
            rank[parent_u]++;
        }
    }
};
class Solution {
public:
    int makeConnected(int n, vector<vector<int>>& connections) {
        Disjoint_set d(n);
        for(auto it:connections){
            d.unionByRank(it[0],it[1]);
        }
        int comp=0;
        for(int i=0;i<d.parent.size();i++){
            if(d.parent[i]==i) comp++;
            cout<<d.parent[i]<<" ";
        }
        cout<<endl;
        int x=0;
        for(auto i:connections){
            if(d.parent[i[0]]==d.parent[i[1]] && d.parent[i[0]]!=i[0]&&d.parent[i[1]]!=i[1]) x++;
        }
        cout<<comp<<" "<<x;
        // if(comp-2<=x) return comp-2;
        // return -1;
         int redundantConnections = connections.size() - (n - comp);

        return redundantConnections >= (comp - 1) ? comp - 2 : -1;
    }
};
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