<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:
    int find_parent(int u,vector<int>&parent){
        if(parent[u] == u){
            return u;
        }
        return parent[u] = find_parent(parent[u],parent);
    }
    bool do_stuff(int u,int v,vector<int>&parent,vector<int>&sizes){
        int p_u = find_parent(u,parent);
        int p_v = find_parent(v,parent);
        if(p_u == p_v){
            return true;
        }
        if(sizes[p_u] < sizes[p_v]){
            parent[p_u] = parent[p_v];
            sizes[p_v] += sizes[p_u];
        }
        else{
            parent[p_v] = parent[p_u];
            sizes[p_u] += sizes[p_v];
        }
        return false;
    }
    void dfs(vector<int>adj[],int u,vector<bool>&vis){
        vis[u] = true;
        for(int v:adj[u]){
            if(vis[v]==false){
                dfs(adj,v,vis);
            }
        }
    }
    int makeConnected(int n, vector<vector<int>>& connections) {
        vector<int>adj[n+1];
        for(int i=0;i<connections.size();i++){
            int u = connections[i][0];
            int v = connections[i][1];
            adj[u].push_back(v);
            adj[v].push_back(u);
        }
        vector<bool>vis(n+1,false);
        int count = 0;
        for(int i=0;i<n;i++){
            if(vis[i]==false){
                dfs(adj,i,vis);
                count++;
            }
        }
        vector<int>parent(n+1,0);
        vector<int>sizes(n+1,1);
        for(int i=0;i<n;i++){
            parent[i] = i;
            sizes[i] = 1;
        }
        int ans = 0;
        for(int i = 0;i<connections.size() && count != 1;i++){
            int u = connections[i][0];
            int v = connections[i][1];
            if(do_stuff(u,v,parent,sizes)==true){
                count--;
                ans++;
            }
        }
        cout << count <<" "<< ans;
        return count == 1 ? ans : -1;
    }
};
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