<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):
        self.groups: Dict[int, int] = {}
        self.size: Dict[int, int] = {}

    def find(self, a: int) -> int:
        if a not in self.groups:
            return -1

        if self.groups[a] != a:
            self.groups[a] = self.find(self.groups[a])

        return self.groups[a]

    def add(self, a: int):
        if a in self.groups:
            return

        self.groups[a] = a
        self.size[a] = 1

    def union(self, a: int, b: int):
        pa, pb = self.find(a), self.find(b)

        if pa == pb:
            return

        if self.size[pa] > self.size[pb]:
            pa, pb = pb, pa

        self.groups[pa] = pb
        self.size[pb] += self.size[pa]


class Solution:
    def _get_prime_factors(self, num: int) -> List[int]:
        factors = []
        curr = 2
        while curr * curr <= num:
            if num % curr == 0:
                num //= curr
                if factors and factors[-1] == curr:
                    continue

                factors.append(curr)
            else:
                curr += 1

        factors.append(num)

        return factors

    def canTraverseAllPairs(self, nums: List[int]) -> bool:
        if len(nums) == 1:
            return True
            
        uf = UnionFind()
        num_lf_map: Dict[int, int] = {}

        for num in nums:
            if num == 1:
                return False

            factors = self._get_prime_factors(num)
            num_lf_map[num] = factors[0]
            uf.add(factors[0])

            for f in factors[1:]:
                uf.add(f)
                uf.union(factors[0], f)

        count = 0
        gid_map = defaultdict(int)
        for num in nums:
            gid_map[uf.find(num_lf_map[num])] += 1

        return len(list(gid_map.keys())) == 1

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