<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:
        
        def helper(num):
            res = []
            while sieve[num] != 0:
                res.append(sieve[num])
                num = num // sieve[num]
            return res

        MAX = max(nums)
        N = len(nums)
        has = [False] * (MAX + 1)
        for x in nums:
            has[x] = True

        # edge cases
        if N == 1:
            return True
        if has[1]:
            return False

        # divisors = [0] * (max(nums) + 1)
        # for num in range(2, max(nums) + 1):
        #     if divisors[num] == 0:
        #         for next_num in range(num, max(nums)+1, num):
        #             divisors[next_num] = num
        sieve = [0] * (MAX + 1)
        for d in range(2, MAX + 1):
            if sieve[d] == 0:
                for v in range(d, MAX + 1, d):
                    sieve[v] = d
        # nums = set(nums)
        # nums = list(nums)
        # for num in nums:
        #     for next_num in helper(num):
        #         self.union(next_num, num)

        # root = self.find(nums[0])
        # for num in nums:
        #     if self.find(num) != root:
        #         return False

        nums = set(nums)
        nums = list(nums)
        self.parent = defaultdict(int)
       
        for i in range(len(nums)):
            next_nums = helper(nums[i])
            print(nums[i], next_nums)
            for j in next_nums:
                self.union(nums[i], j)
        
        #print(self.parent)
        parent = self.find(nums[0])
        for num in nums:
            if self.find(num) != parent:
                #print(num, self.parent[num], parent)
                return False

        return True

        return True

    def find(self, node):

        if node not in self.parent:
            self.parent[node] = node
            return node

        path = []
        while self.parent[node] != node:
            path.append(node)
            node = self.parent[node]

        for p in path:
            self.parent[p] = node

        return node

    def union(self, nodeA, nodeB):
        rootA = self.find(nodeA)
        rootB = self.find(nodeB)

        if rootA != rootB:
            self.parent[rootA] = rootB

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