<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 DisjointSet{
    vector<int> parents;
    vector<int> size;
    public:
    DisjointSet(int n){
        size.resize(n,1);
        parents.resize(n,0);
        for(int i=0;i<n;i++){
            parents[i] = i;
        }
    }
    int getParent(int u){
        if(u==parents[u]) return u;
        return parents[u] = getParent(parents[u]);
    }
    void nodeUnion(int u,int v){
        int upu = parents[u];
        int upv = parents[v];
        if(upu==upv) return;
        if(size[upu]>size[upv]){
            parents[upv] = upu;
            size[upu]+=size[upv];
        }else{
            parents[upu] = upv;
            size[upv]+=size[upu];
        }
    }
};
class Solution {
public:
    void dfs(vector<vector<int>>& graph,int u,vector<bool>& vis){
        vis[u] = true;
        for(auto child:graph[u]){
            if(!vis[child]) dfs(graph,child,vis);
        }
    }
    int totalComponents(int n,vector<vector<int>>& graph){
        vector<bool> vis(n,false);
        int result = 0;
        for(int i=0;i<n;i++){
            if(!vis[i]){
                dfs(graph,i,vis);
                result++;
            }
        }
        return result;
    }
    int makeConnected(int n, vector<vector<int>>& connections) {
        int extracted = 0;
        DisjointSet dj(n);
        vector<vector<int>> graph(n,vector<int>());
        for(auto edge:connections){
            graph[edge[0]].push_back(edge[1]);
            graph[edge[1]].push_back(edge[0]);
            // cout<<"Run";
            int upU = dj.getParent(edge[0]);
            int upV = dj.getParent(edge[1]);
            // cout<<upU<<" :  "<<upV<<endl;
            if(upU!=upV){
                dj.nodeUnion(edge[0],edge[1]);
            }else{
                extracted++;
            }
        }
        int components = totalComponents(n,graph);
        if(components-1>extracted) return -1;
        return components-1;
    }
};
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