<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:
    vector<int> par;
    int findPar(int node){
        if(par[node] == node) return node;
        return par[node] = findPar(par[node]);
    }
    void Union(int u, int v){
        int up = findPar(u);
        int vp = findPar(v);
        par[up] = vp;
    }
    int largestComponentSize(vector<int>& nums) {
        int maxi = *max_element(nums.begin(),nums.end());
        par.resize(maxi+1,0);
        for(int i=1;i<=maxi;i++) par[i] = i;
        for(auto &val : nums){
            for(int k =2;k*k <= val;k++){
                if(val % k == 0){
                    Union(val,k);
                    Union(val,val/k);
                }
            }
        }
        map<int,int> mp;
        for(auto i : nums) mp[findPar(i)]++;
        int res = 0;
        for(auto i : mp) res = max(res,i.second);
        return res;
    }
};
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