<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{
    private:
        vector<int> rank, parent, size;
    public:
        disjointSet(int n){
            rank.resize(n+1);
            size.resize(n+1);
            parent.resize(n+1);
            for(int i=0; i<=n; ++i){
                rank[i]=0;
                size[i]=1;
                parent[i]=i;
            }
        }
        int findUPar(int node){
            if(node==parent[node]){
                return node;
            }else{
                return parent[node]=findUPar(parent[node]);
            }
        }

        void UnionByRank(int u, int v){
            int ULP_u=findUPar(u);
            int ULP_v=findUPar(v);
            if(ULP_u==ULP_v){
                return;
            }
            if(rank[ULP_u]>rank[ULP_v]){
                parent[ULP_v]=ULP_u;
            }
            else if(rank[ULP_u]<rank[ULP_v]){
                parent[ULP_u]=ULP_v;
            }else{
                parent[ULP_u]=ULP_v;
                rank[ULP_u]++;
            }
        }
        void UnionBySize(int u, int v){
            int ULP_u=findUPar(u);
            int ULP_v=findUPar(v);
            if(ULP_u==ULP_v){
                return;
            }
            if(size[ULP_u]>size[ULP_v]){
                parent[ULP_v]=ULP_u;
                size[ULP_u]+=size[ULP_v];
            }
            else{
                parent[ULP_u]=ULP_v;
                size[ULP_v]+=size[ULP_u];
            }
        }
};


class Solution {
public:
    int makeConnected(int n, vector<vector<int>>& connections) {
        disjointSet ds(n);
        int m=connections.size();
        for(auto& it: connections){
            int u=it[0];
            int v=it[1];
            ds.UnionByRank(u, v);
        }
        vector<int> ULP(n, 0);
        for(int i=0; i<n; i++){
            ULP[i]=ds.findUPar(i);
        }
        int cnt=1;
        sort(ULP.begin(), ULP.end());
        for(int i=0; i<n-1 ; i++){
            if(ULP[i]!=ULP[i+1]){
                cnt++;
            }
        }
        if(m>=n-1){
            return cnt-1;
        }
        return -1;
    }
};
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