Problem: Largest Component Size by Common Factor

You are given an integer array of unique positive integers `nums`. Consider
the following graph:

* There are `nums.length` nodes, labeled `nums[0]` to `nums[nums.length - 1]`,
* There is an undirected edge between `nums[i]` and `nums[j]` if `nums[i]` and `nums[j]` share a common factor greater than `1`.

Return the size of the largest connected component in the graph.


Python solution:
import math
import itertools
import bisect
import collections
import string
import heapq
import functools
import sortedcontainers
from typing import List, Dict, Tuple, Iterator

class UnionFind:
  def __init__(self, n: int):
    self.id = list(range(n))
    self.rank = [0] * n

  def unionByRank(self, u: int, v: int) -> None:
    i = self.find(u)
    j = self.find(v)
    if i == j:
      return
    if self.rank[i] < self.rank[j]:
      self.id[i] = j
    elif self.rank[i] > self.rank[j]:
      self.id[j] = i
    else:
      self.id[i] = j
      self.rank[j] += 1

  def find(self, u: int) -> int:
    if self.id[u] != u:
      self.id[u] = self.find(self.id[u])
    return self.id[u]


class Solution:
  def largestComponentSize(self, nums: List[int]) -> int:
    ans = 0
    uf = UnionFind(max(nums) + 1)
    count = collections.Counter()

    for num in nums:
      for x in range(2, int(math.sqrt(num) + 1)):
        if num % x == 0:
          uf.unionByRank(num, x)
          uf.unionByRank(num, num // x)

    for num in nums:
      numRoot = uf.find(num)
      count[numRoot] += 1
      ans = max(ans, count[numRoot])

    return ans


Java solution:
class UnionFind {
  public UnionFind(int n) {
    id = new int[n];
    rank = new int[n];
    for (int i = 0; i < n; ++i)
      id[i] = i;
  }

  public void unionByRank(int u, int v) {
    final int i = find(u);
    final int j = find(v);
    if (i == j)
      return;
    if (rank[i] < rank[j]) {
      id[i] = j;
    } else if (rank[i] > rank[j]) {
      id[j] = i;
    } else {
      id[i] = j;
      ++rank[j];
    }
  }

  public int find(int u) {
    return id[u] == u ? u : (id[u] = find(id[u]));
  }

  private int[] id;
  private int[] rank;
}

class Solution {
  public int largestComponentSize(int[] nums) {
    final int n = Arrays.stream(nums).max().getAsInt();
    int ans = 0;
    UnionFind uf = new UnionFind(n + 1);
    Map<Integer, Integer> count = new HashMap<>();

    for (final int num : nums)
      for (int x = 2; x <= (int) Math.sqrt(num); ++x)
        if (num % x == 0) {
          uf.unionByRank(num, x);
          uf.unionByRank(num, num / x);
        }

    for (final int num : nums)
      ans = Math.max(ans, count.merge(uf.find(num), 1, Integer::sum));

    return ans;
  }
}


C++ solution:
class UnionFind {
 public:
  UnionFind(int n) : id(n), rank(n) {
    iota(id.begin(), id.end(), 0);
  }

  void unionByRank(int u, int v) {
    const int i = find(u);
    const int j = find(v);
    if (i == j)
      return;
    if (rank[i] < rank[j]) {
      id[i] = j;
    } else if (rank[i] > rank[j]) {
      id[j] = i;
    } else {
      id[i] = j;
      ++rank[j];
    }
  }

  int find(int u) {
    return id[u] == u ? u : id[u] = find(id[u]);
  }

 private:
  vector<int> id;
  vector<int> rank;
};

class Solution {
 public:
  int largestComponentSize(vector<int>& nums) {
    const int n = ranges::max(nums);
    int ans = 0;
    UnionFind uf(n + 1);
    unordered_map<int, int> count;

    for (const int num : nums)
      for (int x = 2; x <= sqrt(num); ++x)
        if (num % x == 0) {
          uf.unionByRank(num, x);
          uf.unionByRank(num, num / x);
        }

    for (const int num : nums)
      ans = max(ans, ++count[uf.find(num)]);

    return ans;
  }
};