# 952. Largest Component Size by Common Factor

## Description

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.

Example 1:

Input: nums = [4,6,15,35]
Output: 4


Example 2:

Input: nums = [20,50,9,63]
Output: 2


Example 3:

Input: nums = [2,3,6,7,4,12,21,39]
Output: 8


Constraints:

• 1 <= nums.length <= 2 * 104
• 1 <= nums[i] <= 105
• All the values of nums are unique.

## Solutions

• class UnionFind {
int[] p;

UnionFind(int n) {
p = new int[n];
for (int i = 0; i < n; ++i) {
p[i] = i;
}
}

void union(int a, int b) {
int pa = find(a), pb = find(b);
if (pa != pb) {
p[pa] = pb;
}
}

int find(int x) {
if (p[x] != x) {
p[x] = find(p[x]);
}
return p[x];
}
}

class Solution {
public int largestComponentSize(int[] nums) {
int m = 0;
for (int v : nums) {
m = Math.max(m, v);
}
UnionFind uf = new UnionFind(m + 1);
for (int v : nums) {
int i = 2;
while (i <= v / i) {
if (v % i == 0) {
uf.union(v, i);
uf.union(v, v / i);
}
++i;
}
}
int[] cnt = new int[m + 1];
int ans = 0;
for (int v : nums) {
int t = uf.find(v);
++cnt[t];
ans = Math.max(ans, cnt[t]);
}
return ans;
}
}

• class UnionFind {
public:
vector<int> p;
int n;

UnionFind(int _n)
: n(_n)
, p(_n) {
iota(p.begin(), p.end(), 0);
}

void unite(int a, int b) {
int pa = find(a), pb = find(b);
if (pa != pb) p[pa] = pb;
}

int find(int x) {
if (p[x] != x) p[x] = find(p[x]);
return p[x];
}
};

class Solution {
public:
int largestComponentSize(vector<int>& nums) {
int m = *max_element(nums.begin(), nums.end());
UnionFind* uf = new UnionFind(m + 1);
for (int v : nums) {
int i = 2;
while (i <= v / i) {
if (v % i == 0) {
uf->unite(v, i);
uf->unite(v, v / i);
}
++i;
}
}
vector<int> cnt(m + 1);
int ans = 0;
for (int v : nums) {
int t = uf->find(v);
++cnt[t];
ans = max(ans, cnt[t]);
}
return ans;
}
};

• class UnionFind:
def __init__(self, n):
self.p = list(range(n))

def union(self, a, b):
pa, pb = self.find(a), self.find(b)
if pa != pb:
self.p[pa] = pb

def find(self, x):
if self.p[x] != x:
self.p[x] = self.find(self.p[x])
return self.p[x]

class Solution:
def largestComponentSize(self, nums: List[int]) -> int:
uf = UnionFind(max(nums) + 1)
for v in nums:
i = 2
while i <= v // i:
if v % i == 0:
uf.union(v, i)
uf.union(v, v // i)
i += 1
return max(Counter(uf.find(v) for v in nums).values())


• func largestComponentSize(nums []int) int {
m := slices.Max(nums)
p := make([]int, m+1)
for i := range p {
p[i] = i
}
var find func(int) int
find = func(x int) int {
if p[x] != x {
p[x] = find(p[x])
}
return p[x]
}
union := func(a, b int) {
pa, pb := find(a), find(b)
if pa != pb {
p[pa] = pb
}
}
for _, v := range nums {
i := 2
for i <= v/i {
if v%i == 0 {
union(v, i)
union(v, v/i)
}
i++
}
}
cnt := make([]int, m+1)
for _, v := range nums {
t := find(v)
cnt[t]++
}
return slices.Max(cnt)
}