<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 UnionFind:

    def __init__(self, n):

        self.parent = [node for node in range(n)]
        self.size = [1] * n
        self.components = n
    
    def find(self, node):

        if self.parent[node] == node:
            return node
        
        self.parent[node] = self.find(self.parent[node])
        return self.parent[node]
    
    def union(self, u, v):

        p1, p2 = self.find(u), self.find(v)

        if p1 != p2:

            if self.size[p1] < self.size[p2]:
                self.parent[p1] = p2
                self.size[p2] += self.size[p1]
            
            else:
                self.parent[p2] = p1
                self.size[p1] += self.size[p2]

            self.components -= 1

class Solution:
    def canTraverseAllPairs(self, nums: List[int]) -> bool:

        n = len(nums)

        uf = UnionFind(n)
        factor_index = {}     # f -> index of the value that has factor f.

        for i, n in enumerate(nums):

            f = 2

            while f * f <= n:

                if n % f == 0:

                    if f in factor_index:
                        uf.union(i, factor_index[f])
                    
                    else:
                        factor_index[f] = i

                    while n % f == 0:
                        n //= f
                
                f += 1
                
            if n > 1:

                if n in factor_index:
                    uf.union(factor_index[n], i)
                
                else:
                    factor_index[n] = i
        
        return uf.components == 1

        
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