<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:
        """ basically: for every elem a, sqrt(a) runtime to find all factors of a
        and then dsu.merge to connect all factors...
        once done with all n elems, then run through each elem a and call find(),
        if any of them have different parents, then return false"""
        dsu = DSU(100000)
        for num in nums:
            threshold = int(math.sqrt(num))
            for factor in range(2, threshold + 1):
                if num % factor == 0:
                    dsu.merge(num, factor)
                    dsu.merge(num, num // factor)
        prevNum = 0
        for num in nums:
            if prevNum != 0:
                if dsu.find(prevNum) != dsu.find(num):
                    return False
            if num == 1 and prevNum != 0:
                return False
            prevNum = num
        return True

class DSU:
    def __init__(self, n):
        self.parent = []
        for i in range(0, n + 1):
            self.parent.append(i)
        self.rank = [0] * (n + 1)
        self.numComponents = n
    def find(self, child):
        if self.parent[child] == child:
            return child
        self.parent[child] = self.find(self.parent[child])
        return self.parent[child]
    def merge(self, nodeA, nodeB):
        parA = self.find(nodeA)
        parB = self.find(nodeB)
        if parA != parB:
            if self.rank[parA] == self.rank[parB]:
                self.rank[parA] += 1
                self.parent[parB] = self.parent[parA]
            elif self.rank[parA] > self.rank[parB]:
                self.parent[parB] = self.parent[parA]
            else:
                self.parent[parA] = self.parent[parB]
            self.numComponents -= 1
            return True
        return False

        
5603