<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>

import random
from collections import defaultdict

class Solution:
    def canTraverseAllPairs(self, nums: List[int]) -> bool:
        if len(nums) == 1: return True

        factors_size = int(math.ceil(math.sqrt(100100)))
        
        self.sizes = defaultdict(lambda: 0)
        self.parents = {}
        
        factors = set()
        for i in range(len(nums)):
            num = nums[i]
            if num == 1: return False
            for f in range(2, factors_size):
                if num % f != 0: continue
                factors.add(f)
                self.union(f, nums[i])
                while num % f == 0:
                    num //= f
                if f > num: break

            if num > 1: 
                factors.add(num)
                self.union(num, nums[i])

        ff = list(factors)[0]
            
        return all((self.find(ff) == self.find(f) for f in factors))
  
	
    def find(self, p):
        if p not in self.parents:
            self.parents[p] = p
            return p
        while p != self.parents[p]:
            self.parents[p] = self.parents[self.parents[p]]
            p = self.parents[p]
        return p


    def union(self, p, q):
    
        p_p = self.find(p)
        p_q = self.find(q)
        
        if p_p == p_q: return ####
        
        p_size = self.sizes[p_p]
        q_size = self.sizes[p_q]
        
        if p_size > q_size:
            self.parents[p_q] = p_p
            self.sizes[p_p] += q_size
            
        else:
            self.parents[p_p] = p_q
            self.sizes[p_q] += p_size
            
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