<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 {
public:
    int findUPar(vector<int>& par, int node) {
        if(par[node]==node) return node;
        return par[node]=findUPar(par,par[node]);
    }

    void unionBySize(vector<int>& par, vector<int>& size, int node1, int node2) {
        int ulp1=findUPar(par,node1);
        int ulp2=findUPar(par,node2);
        if(ulp1==ulp2) return;
        if(size[ulp1]<size[ulp2]){
            par[ulp1]=ulp2;
            size[ulp2]+=size[ulp1];
        }
        else if(size[ulp1]>=size[ulp2]){
            par[ulp2]=ulp1;
            size[ulp1]+=size[ulp2];
        }
    }

    int largestComponentSize(vector<int>& nums) {
        vector<int> par;
        vector<int> size(100010,1);
        for(int i=0;i<100010;i++){
            par.push_back(i);
        }

        for(int i=0;i<nums.size();i++){
            for(int f=2;f<=sqrt(nums[i]);f++){
                if(nums[i]%f==0){
                    unionBySize(par,size,nums[i],f);
                    unionBySize(par,size,nums[i],nums[i]/f);
                }
            }
        }

        int res=0;
        unordered_map<int,int> components;
        for(int n:nums){
            int uPar=findUPar(par,n);
            components[uPar]++;
            res=max(res,components[uPar]);
        }

        return res;
    }
};
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