<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 Solution:
    def canTraverseAllPairs(self, nums: List[int]) -> bool:
        N = len(nums)

        parent = [i for i in range(N)]
    
        def ufind(x):
            if parent[x] != x:
                parent[x] = ufind(parent[x])  # Path compression
            return parent[x]
    
        def union(a, b):
            ua = ufind(a)
            ub = ufind(b)
            if ua != ub:
                parent[ua] = ub  # Union without rank for simplicity
    
        Mx = max(nums)
    
        prime = [i for i in range(Mx + 1)]
    
        # Sieve of Eratosthenes to compute smallest prime factors
        for i in range(2, Mx + 1):
            if prime[i] == i:  # i is a prime
                for j in range(i * i, Mx + 1, i):  # Ensure step size is i
                    if prime[j] == j:  # Only update if it's still the original number
                        prime[j] = i
    
        lookup = defaultdict(set)
    
        # Prime factorization and mapping
        for index, x in enumerate(nums):
            while x > 1 and prime[x] != x:
                lookup[prime[x]].add(index)
                x //= prime[x]
            if x > 1:  # Handle last prime factor
                lookup[x].add(index)
    
        # Union sets based on common prime factors
        for k in lookup.keys():
            L = list(lookup[k])
            for i in range(1, len(L)):
                union(L[0], L[i])
    
        # Check if all elements are connected
        root = ufind(0)
        for i in range(1, N):
            if ufind(i) != root:
                return False
        
        return True
        
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