<p>You are given an integer array of unique positive integers <code>nums</code>. Consider the following graph:</p>

<ul>
	<li>There are <code>nums.length</code> nodes, labeled <code>nums[0]</code> to <code>nums[nums.length - 1]</code>,</li>
	<li>There is an undirected edge between <code>nums[i]</code> and <code>nums[j]</code> if <code>nums[i]</code> and <code>nums[j]</code> share a common factor greater than <code>1</code>.</li>
</ul>

<p>Return <em>the size of the largest connected component in the graph</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2018/12/01/ex1.png" style="width: 500px; height: 97px;" />
<pre>
<strong>Input:</strong> nums = [4,6,15,35]
<strong>Output:</strong> 4
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2018/12/01/ex2.png" style="width: 500px; height: 85px;" />
<pre>
<strong>Input:</strong> nums = [20,50,9,63]
<strong>Output:</strong> 2
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2018/12/01/ex3.png" style="width: 500px; height: 260px;" />
<pre>
<strong>Input:</strong> nums = [2,3,6,7,4,12,21,39]
<strong>Output:</strong> 8
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 2 * 10<sup>4</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
	<li>All the values of <code>nums</code> are <strong>unique</strong>.</li>
</ul>

class Solution {
private:
    int findPar(int a,vector<int> &par){
        if(par[a]==a)
        return a;
        par[a]=findPar(par[a],par);
        return par[a];
    }
    void unionPar(int a,int b,vector<int> &par,vector<int> &rank){
        if(rank[a]<=rank[b]){
            par[a]=b;
            rank[b]=max(rank[b],1+rank[a]);
        }else{
            par[b]=a;
            rank[a]=max(rank[a],1+rank[b]);
        }
    }
public:
    int largestComponentSize(vector<int>& nums) {
        unordered_map<int,vector<int>> um;
        for(int i=0;i<nums.size();++i)
        um[nums[i]].push_back(i);
        vector<int> par(nums.size()),rank(nums.size(),1);
        for(int i=0;i<nums.size();++i)
        par[i]=i;
        for(int i=2;i<=100000;++i){
            int fi=-1;
            for(int j=i;j<=100000;j+=i){
                if(um.find(j)!=um.end()){
                    vector<int> v=um[j];
                    if(fi==-1){
                    fi=v[0];
                        for(int k=1;k<v.size();++k){
                            int parA=findPar(fi,par);
                            int parB=findPar(v[k],par);
                            unionPar(parA,parB,par,rank);
                        }
                    }else{
                        for(int k=0;k<v.size();++k){
                            int parA=findPar(fi,par);
                            int parB=findPar(v[k],par);
                            unionPar(parA,parB,par,rank);
                        }
                    }
                }
            }
        }
        unordered_map<int,int> graphCount;
        for(int i=0;i<nums.size();++i)
        graphCount[findPar(i,par)]++;
        int ansr=1;
        for(auto it=graphCount.begin();it!=graphCount.end();++it)
        ansr=max(ansr,it->second);
        return ansr;
    }
};
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