<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 largestComponentSize(vector<int>& nums) {
       
        unordered_map<int,int>num;
        for(auto a:nums)num[a]++;
        unordered_map<int,int>len;
        unordered_map<int,int>parent;
        vector<int>prime(1e5+1,-1);
        if(num[1]>0)len[1]=1;
        for(int i=2;i<=1e5;i++){
            if(prime[i]>0)continue;
            int p=i;
            int c=0;
            for(int x=i;x<=1e5;x+=i){
                if(prime[x]!=-1 && num.find(x)!=num.end()){
                    int pre=p;
                    p=prime[x];
                    if(parent.find(p)==parent.end())parent[p]=p;
                    else {
                        
                        if(parent[p]==p && pre!=p){
                                c+=len[p];
                                len.erase(p);
                                if(pre<p){
                                    parent[p]=pre;
                                    p=pre;
                                }
                                else if(pre>p){
                                    parent[pre]=p;
                                }
                                
                        }
                        else p=parent[p];
                    } 
                }
                prime[x]=p;
                parent[i]=p;
                if(num.find(x)!=num.end() && num[x]==1){
                    c++;
                    num[x]++;
                }

            }
            // if(c>0)cout<<p<<" "<<c<<endl;
            len[p]+=c;
        }
        int ans=0;
        for(auto p:len){
            ans=max(ans,p.second);
        }
        return ans;
    }
};
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