<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>

MAXN = 10**5 + 5
spf = list(range(MAXN))

def sieve():
	spf[1] = 1

	for i in range(4, MAXN, 2):
		spf[i] = 2

	for i in range(3, ceil(sqrt(MAXN))):
		if (spf[i] == i):
			for j in range(i * i, MAXN, i):
				if (spf[j] == j):
					spf[j] = i

def factors(x: int):
	factors_set = set()

	while x != 1:
		factors_set.add(spf[x])
		x //= spf[x]

	return factors_set

sieve()

class Solution:
    def largestComponentSize(self, nums: List[int]) -> int:
        df = {}
        uf = UnionFind(len(nums))
        for pos, num in enumerate(nums):
            for factor in factors(num):
                if factor in df:
                    uf.union(pos, df[factor])
                else:
                    df[factor] = pos
                    
        return uf.mx
            
class UnionFind:
    def __init__(self, n: int):
        self.root_lookup = list(range(n))
        self.rank_lookup = [1] * n
        self.mx = 1

    def find(self, num: int):
        root = self.root_lookup[num]
        while root != self.root_lookup[root]:
              root = self.root_lookup[root]
        self.root_lookup[num] = root
        return root
    
    def union(self, x, y):
        xr, yr = self.find(x), self.find(y)
        if xr == yr:
            return False
        if self.rank_lookup[xr] < self.rank_lookup[yr]:
            xr, yr = yr, xr
        self.root_lookup[yr] = xr
        self.rank_lookup[xr] += self.rank_lookup[yr]
        self.mx = max(self.mx,self.rank_lookup[xr])
        return True
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