Union Find and Prime Factorization

952. Largest Component Size by Common Factor

Union Find is a natural way to combine different sets with connection into one set. So the basic idea is:

• If two numbers have common factor, union them
• During the union process, maintain a set size as the final answer is asking for the set size.

Plain solution

class UF{
    public:
    vector<int> roots;
    vector<int> sizes;
    UF(int n){
        roots.assign(n, 0); for (int i = 0; i < n; ++i)roots[i] = i;
        sizes.assign(n, 1);
}
    int find_set(int x){
        return x == roots[x]? x: roots[x] = find_set(roots[x]);
    }
    void union_set(int x, int y){
        int root_x = find_set(x), root_y = find_set(y);
        if (root_x == root_y)return;
        sizes[root_x] += sizes[root_y];
        roots[root_y] = root_x;
    }
    int get_set_size(int x){
        return sizes[x];
    }
};
class Solution {
    public:
    int largestComponentSize(vector<int>& nums) {
        const int n = nums.size();
        UF uf(n);
        unordered_map<int, vector<int>> graph;
        for (int i = 0; i < n; ++i){
            int num = nums[i];
            for (int factor = 1; factor <= int(sqrt(num)); ++factor){
                if (num % factor == 0){
                    if(factor != 1)graph[factor].push_back(i);
                    if(num / factor != 1)graph[num / factor].push_back(i);
                }
            }
        }
        for (auto [k, v]: graph){
            const int m = v.size();
            for (int i = 1; i < m ; ++i){
                uf.union_set(v[0], v[i]);
            }
        }
        int ret = 0;
        for (int i = 0; i < n; ++i){
            ret = max(ret, uf.get_set_size(i));
        }
        return ret;
    }
};

Speed up by using Prime Factorization

For example, for the number 8 and 12, the have the common factor of

{2, 4}, actually if they have the common prime factor will be enough as non prime factor can be further split to prime factors. In this example, 4 can be split to 2 * 2. So instead of using all factors, we can only use the prime factor to build the connection. For prime factorization, I borrowed it from [1].

class UF{
    public:
    vector<int> roots;
    vector<int> sizes;
    UF(int n){
        roots.assign(n, 0); for (int i = 0; i < n; ++i)roots[i] = i;
        sizes.assign(n, 1);
        
    }
    int find_set(int x){
        return x == roots[x]? x: roots[x] = find_set(roots[x]);
    }
    void union_set(int x, int y){
        int root_x = find_set(x), root_y = find_set(y);
        if (root_x == root_y)return;
        sizes[root_x] += sizes[root_y];
        roots[root_y] = root_x;
    }
    int get_set_size(int x){
        return sizes[x];
    }
};
vector<int> primeFactors(int n)
{
    
    vector<int>ret;
    // Print the number of 2s that divide n
    while (n % 2 == 0)
    {
        ret.push_back(2);
        n = n/2;
    }
 
    // n must be odd at this point. So we can skip
    // one element (Note i = i +2)
    for (int i = 3; i <= sqrt(n); i = i + 2)
    {
        // While i divides n, print i and divide n
        while (n % i == 0)
        {
            ret.push_back(i);
            n = n/i;
        }
    }
 
    // This condition is to handle the case when n
    // is a prime number greater than 2
    if (n > 2)
        ret.push_back(n);
    return ret;
}
class Solution {
public:
    int largestComponentSize(vector<int>& nums) {
        const int n = nums.size();
        UF uf(n);
        unordered_map<int, vector<int>> graph;
        for (int i = 0; i < n; ++i){
            int num = nums[i];
            auto factors = primeFactors(num);
            for (auto &factor: factors)graph[factor].push_back(i);
        }
        for (auto [k, v]: graph){
            const int m = v.size();
            for (int i = 1; i < m ; ++i){
                    uf.union_set(v[0], v[i]);
            }
        }
        int ret = 0;
        for (int i = 0; i < n; ++i){
            ret = max(ret, uf.get_set_size(i));
        }
        return ret;
    }
};

Run time

Time difference Top submission is the plain version and the bottom one is using the Prime Factorization version

Reference

[1] https://www.geeksforgeeks.org/print-all-prime-factors-of-a-given-number/