<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 math import sqrt

class Solution:
    def canTraverseAllPairs(self, nums: List[int]) -> bool:
        maxNum = max(nums)
        upper =  maxNum # int(sqrt(maxNum))
        primeToNums = defaultdict(set)
        numSet = dict()



        parent = [i for i,_ in enumerate(nums)]
        size = [1 for _ in nums]

        def find(x):
            if parent[x] == x:
                return x
            
            parent[x] = find(parent[x])
            return parent[x]

        def union(a, b):
            a = find(a)
            b = find(b)
            if a == b:
                return

            if size[a] < size[b]:
                parent[a] = b
                size[b] += size[a]
            else:
                parent[b] = a
                size[a] += size[b]

        for i, num in enumerate(nums):
            if num in numSet:
                if num == 1:
                    return False
                union(i, numSet[num])
            numSet[num] = i

        def findPrimes(maxNum):
            primes = set(range(2, maxNum + 1))
            for num in range(2, maxNum + 1):
                if num not in primes:
                    continue

                if num in numSet:
                    primeToNums[num].add(numSet[num])
                
                for i in range(num * 2, maxNum + 1, num):
                    if i in primes:
                        primes.remove(i)
                    if i in numSet:
                        primeToNums[num].add(numSet[i])
                    
            return primes

        primes = findPrimes(maxNum)
        # print(primeToNums)
        # print(findPrimes(maxNum))
                


        def mergeByDiv(div):
            curNums = []
            for idx, num in enumerate(nums):
                if num % div == 0:
                    curNums.append(idx)

            if curNums:
                root = curNums[0]

                for i in range(1, len(curNums)):
                    union(root, curNums[i])

        for prime in primes:
            l = list(primeToNums[prime])
            if l:
                root = l[0]
                for i in range(1, len(l)):
                    union(root, l[i])
        # mergeByDiv(2)
        # for i in primes:
        #     mergeByDiv(i)

        return size[find(0)] == len(nums)
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