<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 DisSet{
    public:
    vector<int> parent,size;
    DisSet(int n){
        size.resize(n+1,1);
        parent.resize(n+1);
        for(int i=0;i<=n;i++){
            parent[i]=i;
        }
    }
    int findParent(int node){
        if(parent[node]==node) return node;
        return parent[node]=findParent(parent[node]);
    }
    void unionOfNodes(int a ,int b){
        int upa=findParent(a);
        int upb=findParent(b);
        if(size[a]>size[b]){
            size[a]+=size[b];
            parent[upb]=upa;
        }
        else{
            size[b]+=size[a];
            parent[upa]=upb;
        }
    }
};
class Solution {
public:

    // void bfs(int node,vector<int> adj[],vector<int> &vis){
    //     vis[node]=1;
    //     queue<int> q;
    //     q.push(node);
    //     while(!q.empty()){
    //         int nd=q.front();
    //         q.pop();
    //         for(auto &it:adj[nd]){
    //             if(!vis[it]){
    //                 vis[it]=1;
    //                 q.push(it);
    //             }
    //         }
    //     }
    // }

    int makeConnected(int n, vector<vector<int>>& connections) {
        if(connections.size()<n-1) return -1;
        DisSet ds(n);
        for(auto &it:connections){
            if(ds.findParent(it[0])!=ds.findParent(it[1])){
                ds.unionOfNodes(it[0],it[1]);
            }
        }
        int ans=0;
        for(int i=0;i<n;i++){
            if(ds.parent[i]==i) ans++;
        }
        return ans-1;
    }
};
const static auto fast = []
{
    std::ios_base::sync_with_stdio(false);
    std::cin.tie(nullptr);
    std::cout.tie(nullptr);
    return 0;
}();
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