<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 disjointSet {
        vector<int> rank;
        vector<int> parent;
        public:

        disjointSet(int n) {
            rank = vector<int> (n, 0);
            parent = vector<int> (n, 1);

            for(int i=0;i<n;i++)
                parent[i] = i;
        }

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

        void Union(int x, int y) {
            int x_rep = find(x), y_rep = find(y);
            if(x_rep == y_rep)
                return;
            
            if(rank[x_rep] < rank[y_rep])
                parent[x_rep] = y_rep;
            else if(rank[y_rep] < rank[x_rep])
                parent[y_rep] = x_rep;
            else {
                parent[y_rep] = x_rep;
                rank[x_rep]++;
            }
        }

        int getComponents() {
            int cnt = 0;
            for(int i=0;i<parent.size();i++) {
                if(parent[i] == i)
                    cnt++;
            }
            return cnt;
        }
        
    };

    int makeConnected(int n, vector<vector<int>>& connections) {
        
        disjointSet *ds = new disjointSet(n);

        int cnt = 0;
        for(auto it : connections) {
            if(ds->find(it[0]) != ds->find(it[1])) {
                ds->Union(it[0], it[1]);
            }
            else {
                cnt++;
            }
        }

        vector<vector<int>> adj(n);
        for(auto it : connections) {
            adj[it[0]].push_back(it[1]);
            adj[it[1]].push_back(it[0]);
        }

        int comp = ds->getComponents();
        if(comp-1 <= cnt)
            return comp - 1;
        
        return -1;
    }
};
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