<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{
        public:
            vector<int> rank, parent;
            
            Disjointset(int n){
                rank.resize(n, 0);
                parent.resize(n);

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

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

                return parent[x] = Find(parent[x]);
            }

            void UnionbyRank(int x, int y){
                int ulx = Find(x);
                int uly = Find(y);

                if(ulx == uly) return;
                if(rank[ulx] < rank[uly]){
                    parent[ulx] = uly;
                }
                else if(rank[uly] < rank[ulx]){
                    parent[uly] = ulx;
                }
                else{
                    parent[ulx] = uly;
                    rank[uly]++;
                }
            }
    };

    int makeConnected(int n, vector<vector<int>>& connections) {
        Disjointset ds(n);
        int extra = 0;

        for(auto it:connections){
            int u = it[0];
            int v = it[1];

            if(ds.Find(u) == ds.Find(v)){
                extra++;
            }
            else ds.UnionbyRank(u, v);
        }

        int cntC = 0;
        for(int i = 0; i<n; i++){
            if(ds.parent[i] == i)cntC++;
        }

        int req = cntC-1;
        if(extra >= req) return req;
        return -1;
    }
};
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