Rated Groups

A group of N people has some distinct integer ratings 'R' and a consistency factor 'C'. People who frequently participate have consistency factor 1
and those who rarely participate have a consistency factor 0

People create groups by following all these conditions:

1. Ri, Rj are ratings of two-person. If Ri is divisible by Rj, jth person joins ith person's group.
2.Consistency factor of teammates in a group must be the same.
3.No Group must be a proper subset of other groups.
4.A person may be a member of more than one group.

Your task is to find the numbers of groups created.

Example:

Input:  1
5
8 6 2 3 4
1 1 1 0 1
Output: 3

Approach

Java

import java.util.ArrayList;
import java.util.Collections;
import java.util.HashSet;

public class RatedGroups {
    public static void main(String args[]) throws Exception {
        int n = 5;
        int arr[] = { 8, 6, 2, 3, 4 };
        int arr2[] = { 1, 1, 1, 0, 1 };

        int freqo[] = new int[10001];
        int freqz[] = new int[10001];
        boolean visitedo[] = new boolean[10001];
        boolean visitedz[] = new boolean[10001];

        HashSet<Integer> setone = new HashSet<Integer>();
        HashSet<Integer> setzero = new HashSet<Integer>();
        ArrayList<Integer> one = new ArrayList<Integer>();
        ArrayList<Integer> zero = new ArrayList<Integer>();

        int temp = 0;
        for (int i = 0; i < n; i++) {
            temp = arr2[i];
            if (temp == 1) {
                one.add(arr[i]);// storing in list one
                freqo[arr[i]]++; // increasing the frequecny
                setone.add(arr[i]);// adding to set so that we can find it faster
            }

            else {
                zero.add(arr[i]);// storing in list zero
                freqz[arr[i]]++; // increasing the frequency
                setzero.add(arr[i]);// so we can find it faster
            }

        }
        Collections.sort(one, Collections.reverseOrder());
        Collections.sort(zero, Collections.reverseOrder());
        int count = 0;
        // solving the question for consistency = 1
        for (Integer i : one) {
            if (visitedo[i])
                continue;
            visitedo[i] = true;
            count += freqo[i];
            for (int j = 1; j * j <= i; j++) {
                if (i % j == 0) {
                    if (setone.contains(j))
                        visitedo[j] = true;
                    if (i / j != j && setone.contains(i / j))
                        visitedo[i / j] = true;
                }
            }
        }

        // solving the question for consistency = 1
        for (Integer i : zero) {
            if (visitedz[i])
                continue;
            visitedz[i] = true;
            count += freqz[i];
            for (int j = 1; j * j <= i; j++) {
                if (i % j == 0) {
                    if (setzero.contains(j))
                        visitedz[j] = true;
                    if (i / j != j && setzero.contains(i / j))
                        visitedz[i / j] = true;
                }
            }
        }
        System.out.println(count);
    }
}