<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>

from typing import List
from collections import defaultdict

class UnionFind:
    def __init__(self, n):
        self.parent = list(range(n))
        self.rank = [1] * n  # Rank for union by rank

    def find(self, x):
        # Path compression
        if self.parent[x] != x:
            self.parent[x] = self.find(self.parent[x])
        return self.parent[x]

    def union(self, x, y):
        # Union by rank
        rootX, rootY = self.find(x), self.find(y)
        if rootX != rootY:
            if self.rank[rootX] > self.rank[rootY]:
                self.parent[rootY] = rootX
            elif self.rank[rootX] < self.rank[rootY]:
                self.parent[rootX] = rootY
            else:
                self.parent[rootY] = rootX
                self.rank[rootX] += 1


class Solution:
    def canTraverseAllPairs(self, nums: List[int]) -> bool:
        if len(nums) == 1:
            return True
        
        l = len(nums)
        uf = UnionFind(l)
        m = max(nums)
        self.memo = {}

        def sieve_of_eratosthenes(n):
            prime = [True] * (n + 1)
            prime[0] = prime[1] = False
            for p in range(2, int(n**0.5) + 1):
                if prime[p]:
                    for i in range(p * p, n + 1, p):
                        prime[i] = False
            return [p for p in range(2, n + 1) if prime[p]]

        primes = sieve_of_eratosthenes(m)

        def prime_factors(n):
            if n in self.memo:
                return self.memo[n]
            original_n = n
            factors = []
            for prime in primes:
                if prime * prime > n:
                    break
                if n % prime == 0:
                    factors.append(prime)
                while n % prime == 0:
                    n //= prime
                    if n in self.memo:
                        factors += self.memo[n]
                        self.memo[original_n] = factors
                        return factors
            if n > 1:
                factors.append(n)
            self.memo[original_n] = factors
            return factors

        d = defaultdict(int)  # Dictionary to store prime factors and their indices
        for i in range(l):
            if nums[i] == 1:
                return False  # If any number is 1, return False
            factors = prime_factors(nums[i])
            for factor in factors:
                if d[factor] == 0:
                    d[factor] = i + 1  # Avoid zero-based index issues
                else:
                    uf.union(i, d[factor] - 1)
                    d[factor] = i + 1

        # Check if all elements are connected
        root = uf.find(0)
        for i in range(1, l):
            if uf.find(i) != root:
                return False
        return True

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