<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:
    int find( int x , vector<int>& parent )
    {
        if( parent[x] == x ) return x;
        return parent[x] = find( parent[x] , parent );
    }

    void unionbyrank( int a, int b, vector<int>& parent)
    {
            int x = find( a , parent );
            int y = find( b , parent );
            parent[x] = y;
    }

vector<int> f(int N)
{
    vector<int>ans;
    if (N == 2) {
        return{2};
    }
 
    unordered_map<int, bool> visited;
 
    for (int i = 2; i * i <= N; i++) {
        while (N % i == 0) {
            if (!visited[i]) {
                ans.push_back( i );
                // cout << i << " ";
                visited[i] = 1;
            }
            N /= i;
        }
    }
    if (N > 2)
        ans.push_back( N );

    return ans;
}
    bool canTraverseAllPairs(vector<int>& nums) {
        sort( nums.begin() , nums.end() );
        int n = nums.size();
        if( nums[0] == 1 ) 
            return n == 1 ? 1 : 0;
        vector<int>parent(n);
        for( int i=0;i<n;i++ )
            parent[i] = i;
        vector<int>prime( 1e5 , -1 );
        for( int i=0;i<n;i++ )
        {
            int num = nums[i];
            vector<int>temp = f( num );
            for( int fac : temp )
            {
                if( prime[fac] != -1 )//prime[fac] is an index
                    unionbyrank( prime[fac], i , parent );//i becomes parent
                
                prime[fac] = i;
            }
        }
        int cnt = 0;    
        for( int i=0;i<n;i++ )
           { if( parent[i] == i ) cnt++;
            if( cnt > 1 ) return 0;
           }

        return cnt == 1 ? 1 : 0;
    }
};
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