<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, size):
        self.parent = list(range(size))
        self.rank = [1] * size

    def find(self, u):
        if self.parent[u] != u:
            self.parent[u] = self.find(self.parent[u])
        return self.parent[u]

    def union(self, u, v):
        rootU = self.find(u)
        rootV = self.find(v)
        if rootU != rootV:
            if self.rank[rootU] > self.rank[rootV]:
                self.parent[rootV] = rootU
            elif self.rank[rootU] < self.rank[rootV]:
                self.parent[rootU] = rootV
            else:
                self.parent[rootV] = rootU
                self.rank[rootU] += 1

def get_prime_factors(n):
    factors = set()
    if n % 2 == 0:
        factors.add(2)
        while n % 2 == 0:
            n //= 2
    for i in range(3, int(n**0.5) + 1, 2):
        if n % i == 0:
            factors.add(i)
            while n % i == 0:
                n //= i
    if n > 2:
        factors.add(n)
    return factors

class Solution:
    def canTraverseAllPairs(self, nums: List[int]) -> bool:
        n = len(nums)
        uf = UnionFind(n)
        
        prime_to_index = defaultdict(list)
        
        for i, num in enumerate(nums):
            prime_factors = get_prime_factors(num)
            for prime in prime_factors:
                prime_to_index[prime].append(i)
        
        for indices in prime_to_index.values():
            for i in range(1, len(indices)):
                uf.union(indices[0], indices[i])
        
        root = uf.find(0)
        return all(uf.find(i) == root for i in range(n))
1941