<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 UnionFinder:
    def __init__(self, n):
        self.parents = list(range(n))
        self.ranks = [0] * n
    
    def union(self, l, r):
        r_l = self.find(l)
        r_r = self.find(r)

        if r_l == r_r:
            return True
        if self.ranks[r_l] > self.ranks[r_r]:
            self.parents[r_r] = r_l
        elif self.ranks[r_l] < self.ranks[r_r]:
            self.parents[r_l] = r_r
        else:
            self.ranks[r_l] += 1
            self.parents[r_r] = r_l
    
    def find(self, l):
        if self.parents[l] == l:
            return l
        else:
            root = self.find(self.parents[l])
            self.parents[l] = root
            return root


class Solution:
    def canTraverseAllPairs(self, nums: List[int]) -> bool:
        if len(nums) == 1:
            return True
        nums_set = set(nums)
        if 1 in nums_set:
            return False
        nums = list(nums)
        prime_set_list =[]
        all_primes = set([])
        for num in nums:
            prime_set = self.get_prime_set(num)
            all_primes.update(prime_set)
            prime_set_list.append(prime_set)

        all_primes = list(all_primes)
        indices = {}
        for i, p in enumerate(all_primes):
            indices[p] = i

        print(all_primes)
        print(prime_set_list)

        self.finder = UnionFinder(len(all_primes))

        for prime_set in prime_set_list:
            prime_list = list(prime_set)
            if not prime_list:
                continue
            p0 = prime_list[0]
            for p in prime_list[1:]:
                self.finder.union(indices[p0], indices[p])
        
        root_set = set([])
        for i in range(len(all_primes)):
            root_set.add(self.finder.find(i))
        return len(root_set) == 1

    
    def get_prime_set(self, num):
        prime = 2
        prime_set = set([])
        while prime**2 <= num:
            while num % prime == 0:
                prime_set.add(prime)
                num //= prime
            prime += 1
        if num != 1:
            prime_set.add(num)
        return list(prime_set)


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