<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>

import math
class Solution:
    def largestComponentSize(self, nums: List[int]) -> int:
        uf = UnionFind(nums)
        for x in nums:
            for f in range(1, int(math.sqrt(x)) + 1):
                if x % f == 0 and f > 1:
                    uf.add(f)
                    uf.union(x, f)
                if x % f == 0 and x // f > 1:
                    uf.add(x // f)
                    uf.union(x, x // f)
        return max(list(uf.div_size.values()))

class UnionFind:
    def __init__(self, nums: List[int]) -> None:
        self.div_size = {}
        # used to denote divider's parent
        self.div_parent = {}
        # used to denote num's parent
        self.num_parent = {x: -1 for x in nums}
    
    # find the parent which must be divider
    def find(self, a: int, is_div: bool) -> int:
        if not is_div:
            a = self.num_parent[a]
            if a == -1:
                return -1
        if self.div_parent[a] != a:
            self.div_parent[a] = self.find(self.div_parent[a], True)
        return self.div_parent[a]
    
    # order matters in this problem, b is in cluster, a is not in cluster
    def union(self, a: int, b: int) -> None:
        pa = self.find(a, False)
        pb = self.find(b, True)
        if pa != pb:
            if pa == -1:
                self.div_size[pb] += 1
                self.num_parent[a] = pb
            else:
                self.div_size[pb] += self.div_size[pa]
                self.div_parent[pa] = pb

    def add(self, a: int) -> None:
        if a not in self.div_parent:
            self.div_size[a] = 0
            self.div_parent[a] = a
24950