<p>You are given an integer array of unique positive integers <code>nums</code>. Consider the following graph:</p>

<ul>
	<li>There are <code>nums.length</code> nodes, labeled <code>nums[0]</code> to <code>nums[nums.length - 1]</code>,</li>
	<li>There is an undirected edge between <code>nums[i]</code> and <code>nums[j]</code> if <code>nums[i]</code> and <code>nums[j]</code> share a common factor greater than <code>1</code>.</li>
</ul>

<p>Return <em>the size of the largest connected component in the graph</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2018/12/01/ex1.png" style="width: 500px; height: 97px;" />
<pre>
<strong>Input:</strong> nums = [4,6,15,35]
<strong>Output:</strong> 4
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2018/12/01/ex2.png" style="width: 500px; height: 85px;" />
<pre>
<strong>Input:</strong> nums = [20,50,9,63]
<strong>Output:</strong> 2
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2018/12/01/ex3.png" style="width: 500px; height: 260px;" />
<pre>
<strong>Input:</strong> nums = [2,3,6,7,4,12,21,39]
<strong>Output:</strong> 8
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 2 * 10<sup>4</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
	<li>All the values of <code>nums</code> are <strong>unique</strong>.</li>
</ul>

class UnionFind:
    def __init__(self, size):
        self.parent = [i for i in range(size)]
    
    def find(self, x):
        while self.parent[x] != x:
            # Path compression
            self.parent[x] = self.parent[self.parent[x]]
            x = self.parent[x]
        return x
    
    def union(self, x, y):
        px, py = self.find(x), self.find(y)
        if px != py:
            self.parent[py] = px

class Solution:
    def largestComponentSize(self, nums):
        n = len(nums)
        uf = UnionFind(n)
        prime_to_index = {}
        
        for i, num in enumerate(nums):
            primes = self.get_prime_factors(num)
            for p in primes:
                if p in prime_to_index:
                    uf.union(i, prime_to_index[p])
                else:
                    prime_to_index[p] = i
        
        component_size = {}
        max_size = 0
        for i in range(n):
            root = uf.find(i)
            component_size[root] = component_size.get(root, 0) + 1
            max_size = max(max_size, component_size[root])
        
        return max_size
    
    def get_prime_factors(self, n):
        factors = set()
        # Divide by 2 until n is odd
        while n % 2 == 0:
            factors.add(2)
            n //= 2
        # Check for odd factors
        p = 3
        while p * p <= n:
            while n % p == 0:
                factors.add(p)
                n //= p
            p += 2
        if n > 1:
            factors.add(n)
        return factors

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