<p>You are given a <strong>0-indexed</strong> integer array <code>nums</code>, and you are allowed to <strong>traverse</strong> between its indices. You can traverse between index <code>i</code> and index <code>j</code>, <code>i != j</code>, if and only if <code>gcd(nums[i], nums[j]) &gt; 1</code>, where <code>gcd</code> is the <strong>greatest common divisor</strong>.</p>

<p>Your task is to determine if for <strong>every pair</strong> of indices <code>i</code> and <code>j</code> in nums, where <code>i &lt; j</code>, there exists a <strong>sequence of traversals</strong> that can take us from <code>i</code> to <code>j</code>.</p>

<p>Return <code>true</code><em> if it is possible to traverse between all such pairs of indices,</em><em> or </em><code>false</code><em> otherwise.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [2,3,6]
<strong>Output:</strong> true
<strong>Explanation:</strong> In this example, there are 3 possible pairs of indices: (0, 1), (0, 2), and (1, 2).
To go from index 0 to index 1, we can use the sequence of traversals 0 -&gt; 2 -&gt; 1, where we move from index 0 to index 2 because gcd(nums[0], nums[2]) = gcd(2, 6) = 2 &gt; 1, and then move from index 2 to index 1 because gcd(nums[2], nums[1]) = gcd(6, 3) = 3 &gt; 1.
To go from index 0 to index 2, we can just go directly because gcd(nums[0], nums[2]) = gcd(2, 6) = 2 &gt; 1. Likewise, to go from index 1 to index 2, we can just go directly because gcd(nums[1], nums[2]) = gcd(3, 6) = 3 &gt; 1.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [3,9,5]
<strong>Output:</strong> false
<strong>Explanation:</strong> No sequence of traversals can take us from index 0 to index 2 in this example. So, we return false.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [4,3,12,8]
<strong>Output:</strong> true
<strong>Explanation:</strong> There are 6 possible pairs of indices to traverse between: (0, 1), (0, 2), (0, 3), (1, 2), (1, 3), and (2, 3). A valid sequence of traversals exists for each pair, so we return true.
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
</ul>

class Solution {
public:
  void getPrimeFactors(set<int>&s , int x) {
        vector<int> primeFactors;
        for (int i = 2; i * i <= x; i++) {
            if (x % i == 0) {
                s.insert(i);
                while (x % i == 0) {
                    x /= i;
                }
            }
        }
        if (x != 1) {
            s.insert(x);
        }
    }

    int findparent(unordered_map<int,int>&parent , int x)
    {
        if(parent[x] == x)
           return x;
        return findparent(parent , parent[x]);   
    }

    bool canTraverseAllPairs(vector<int>& nums) {

        if(nums.size()==1)
           return true;
      unordered_map<int,int>parent;
      unordered_map<int,int>rank;
      for(int i=0;i<nums.size();i++)
      {
        if(nums[i] == 1)
           return false;
           
         set<int>s;
         getPrimeFactors(s , nums[i]);
     //    cout<<nums[i]<<endl;
     //    cout<<"factors = ";
         for(auto it1 =s.begin() ;it1 != s.end();it1++)
         {
    //        cout<<*it1<<" ";
           parent[*it1] = *it1;
           rank[*it1] = 1;
         }
         parent[nums[i]] = nums[i];
         rank[nums[i]] = 1;
    //     cout<<endl;
      }   

      for(auto it = parent.begin() ; it!=parent.end() ;it++)
      {
         set<int>s;
         getPrimeFactors(s , it->first);

         for(auto it1 =s.begin() ;it1 != s.end();it1++)
         {
            if(parent[*it1] != 0)
            {
                int parent1 = findparent(parent , it->first);
                int parent2 = findparent(parent , *it1);
                
                if(parent1 != parent2)
                {
                if(rank[parent1] > rank[parent2])
                   parent[parent2] = parent1; 
                else if(rank[parent2] > rank[parent1])
                   parent[parent1] = parent2;
                else 
                {
                    parent[parent2] = parent1;
                    rank[parent1]+=1;
                } 
                }   
            }
         }
      }
         
         set<int>p;
         for(auto it=parent.begin() ;it!= parent.end() ;it++)
         {
            p.insert(findparent(parent , it->first));
         }

         return p.size() == 1 ;

      }  
};
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