<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:
    vector<int> rank;
    vector<int> par;
    int find(int x) {
        if(x == par[x]) {
            return x;
        }
        par[x] = find(par[x]);
        return par[x];
    }
    int union_by_rank(int x, int y) {
        x = find(x);
        y = find(y);
        if(x == y)
        return 1;

        if(rank[x] < rank[y]) {
            rank[x] = rank[y];
            par[x] = y;
        } else if(rank[x] > rank[y]) {
            rank[y] = rank[x];
            par[y] = x;
        } else {
            rank[x]++;
            rank[y] = rank[x];
            par[y] = x;
        }
        return 0;
    }
    int makeConnected(int n, vector<vector<int>>& connections) {
        int i;
        for(i=0;i<n;i++) {
            rank.push_back(0);
            par.push_back(i);
        }

        set<int> st;
        int cnt = 0;
        for(auto edge: connections) {
            cnt += union_by_rank(edge[0], edge[1]);
        }

        for(i=0;i<n;i++) {
            if(par[i] == i)
            st.insert(i);
        }

        return cnt >= st.size() - 1 ? st.size() - 1: -1;

    }
};
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