<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: int):
    self.id = list(range(n))
    self.sz = [1] * n

  def unionBySize(self, u: int, v: int) -> None:
    i = self._find(u)
    j = self._find(v)
    if i == j:
      return
    if self.sz[i] < self.sz[j]:
      self.sz[j] += self.sz[i]
      self.id[i] = j
    else:
      self.sz[i] += self.sz[j]
      self.id[j] = i

  def getSize(self, i: int) -> int:
    return self.sz[i]

  def _find(self, u: int) -> int:
    if self.id[u] != u:
      self.id[u] = self._find(self.id[u])
    return self.id[u]


class Solution:
  def canTraverseAllPairs(self, nums: List[int]) -> bool:
    n = len(nums)
    max_num = max(nums)
    maxPrimeFactor = self._sieveEratosthenes(max_num + 1)
    primeToFirstIndex = collections.defaultdict(int)
    uf = UnionFind(n)

    for i, num in enumerate(nums):
      for prime_factor in self._getPrimeFactors(num, maxPrimeFactor):
        if prime_factor in primeToFirstIndex:
          uf.unionBySize(primeToFirstIndex[prime_factor], i)
        else:
          primeToFirstIndex[prime_factor] = i

    return any(uf.getSize(i) == n for i in range(n))

  def _sieveEratosthenes(self, n: int) -> List[int]:
    """Gets the minimum prime factor of i, where 1 < i <= n."""
    minPrimeFactors = [i for i in range(n + 1)]
    for i in range(2, int(n**0.5) + 1):
      if minPrimeFactors[i] == i:  # `i` is prime.
        for j in range(i * i, n, i):
          minPrimeFactors[j] = min(minPrimeFactors[j], i)
    return minPrimeFactors

  def _getPrimeFactors(self, num: int, minPrimeFactors: List[int]) -> List[int]:
    primeFactors = []
    while num > 1:
      divisor = minPrimeFactors[num]
      primeFactors.append(divisor)
      while num % divisor == 0:
        num //= divisor
    return primeFactors
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