<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 findParent(unordered_map<int, int> &parent, int a)
    {
        if(parent[a] != a)
        {
            return findParent(parent, parent[a]);
        }

        return parent[a];
    }


    int make_component(unordered_map<int,int> &rank, unordered_map<int,int>& parent, int a, int b)
    { 
        int aParent = findParent(parent, a);
        int bParent = findParent(parent, b);

        if(aParent != bParent)
        {
            if(rank[aParent] < rank[bParent])
            {
                parent[aParent] = bParent;
                rank[bParent] += rank[aParent];
                return rank[bParent];
            }
            else if(rank[bParent] < rank[aParent])
            {
                parent[bParent] = aParent;
                rank[aParent] += rank[bParent];
                return rank[aParent];                
            }
            else
            {
                parent[aParent] = bParent;
                rank[bParent]+=rank[aParent];
                return rank[bParent];
            }

        }

        else {
            return rank[aParent];
        }
        return 0;
    }


    int largestComponentSize(vector<int>& nums) {

        unordered_map<int,int> rank;

        unordered_map<int, int> parent;

        for(auto i : nums)
        {
            parent[i] = i;
            rank[i] = 1;
        }

        int ans = 1;

        unordered_map<int, int> mp;

        for(int i = 0; i < nums.size(); i++)
        {
            int num = nums[i];

            for(int p = 2; p*p <= num; p++)
            {
                if(num%p == 0)
                {
                    if(mp.find(p) == mp.end())
                    {
                        mp[p] = num;
                    }
                    else 
                    {
                        // connect these nodes
                        ans = max(ans,make_component(rank, parent, num,mp[p]));
                    }

                    if(mp.find(num/p) == mp.end())
                    {
                        mp[num/p] = num;
                    }
                    else
                    {
                        ans = max(ans,make_component(rank, parent, num, mp[num/p]));
                    }
                }
            }

            if(mp.find(num) == mp.end())
            {
                mp[num] = num;
            }
            else
            {
                ans = max(ans, make_component(rank, parent, mp[num], num));
            }
        }

        return ans;
    }
};
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