Hard
Largest Component Size by Common Factor — C++
Full explanation · Time O(f * n) · Space O(p + n)
// Time: O(f * n), f is the max number of unique prime factors
// Soace: O(p + n), p is the total number of unique primes
class Solution {
public:
int largestComponentSize(vector<int>& A) {
UnionFind union_find(A.size());
unordered_map<int, int> nodesWithCommonFactor;
for (int i = 0; i < A.size(); ++i) {
for (const auto& factor : primeFactors(A[i])) {
if (!nodesWithCommonFactor.count(factor)) {
nodesWithCommonFactor[factor] = i;
}
union_find.union_set(nodesWithCommonFactor[factor], i);
}
}
return union_find.max_size();
}
private:
vector<int> primeFactors(int i) const {
vector<int> result;
int d = 2;
if (i % d == 0) {
while (i % d == 0) {
i /= d;
}
result.emplace_back(d);
}
d = 3;
for (d = 3; d * d <= i; d += 2) {
if (i % d == 0) {
while (i % d == 0) {
i /= d;
}
result.emplace_back(d);
}
}
if (i != 1) {
result.emplace_back(i);
}
return result;
}
class UnionFind {
public:
UnionFind(const int n) : set_(n), size_(n, 1) {
iota(set_.begin(), set_.end(), 0);
}
int find_set(const int x) {
if (set_[x] != x) {
set_[x] = find_set(set_[x]); // Path compression.
}
return set_[x];
}
bool union_set(const int x, const int y) {
int x_root = find_set(x), y_root = find_set(y);
if (x_root == y_root) {
return false;
}
set_[min(x_root, y_root)] = max(x_root, y_root);
size_[max(x_root, y_root)] += size_[min(x_root, y_root)];
return true;
}
int max_size() const {
return *max_element(size_.cbegin(), size_.cend());
}
private:
vector<int> set_;
vector<int> size_;
};
};