<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>

from math import sqrt, ceil
from functools import cache

class Solution:
    def largestComponentSize(self, nums: List[int]) -> int:
        def factor(n):
            if n in factors:
                return factors[n]
            
            ret = set()
            sqroot = sqrt(n)
            for i in range(2, ceil(sqroot)+1):
                if n % i == 0:
                    ret |= factor(n//i)
                    ret |= factor(i)
                    break
            if len(ret) == 0:
                ret = {n}
            factors[n] = ret
            return ret
        
        def find(x):
            parents.setdefault(x, x)
            sizes.setdefault(x, 1)
            if parents[x] != x:
                parents[x] = find(parents[x])
            return parents[x]

        def add_union(x, y):
            nonlocal max_size

            rootX = find(x)
            rootY = find(y)
            if rootX != rootY:
                sizeX = sizes[rootX]
                sizeY = sizes[rootY]
                if sizeX > sizeY:
                    parents[rootY] = rootX
                    sizes[rootX] += sizes[rootY]
                    max_size = max(max_size, sizes[rootX])
                else:
                    parents[rootX] = rootY
                    sizes[rootY] += sizes[rootX]
                    max_size = max(max_size, sizes[rootY])


        factors = {1: set(), 2: {2}}
        parents = {}
        # ranks = {}
        sizes = {}
        max_size = 1

        # print(f"{parents=}")
        # print(f"{sizes=}")
        # print(f"{factors=}")
        for n in nums:
            for f in factor(n):
                add_union(n, f)
                # print(f"Added union: ({n}, {f})")
                # print(f"\t{parents=}")
                # print(f"\t{sizes=}")
                # print(f"\t{factors=}")
        counts = defaultdict(int)
        for n in nums:
            counts[find(n)] += 1
        return max(counts.values())
50939