Between Two Sets in Algorithm | HackerRank Programming Solutions | HackerRank Problem Solving Solutions in Java [💯Correct]

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Introduction To Algorithm

The word Algorithm means “a process or set of rules to be followed in calculations or other problem-solving operations”. Therefore Algorithm refers to a set of rules/instructions that step-by-step define how a work is to be executed upon in order to get the expected results. 

Advantages of Algorithms:

  • It is easy to understand.
  • Algorithm is a step-wise representation of a solution to a given problem.
  • In Algorithm the problem is broken down into smaller pieces or steps hence, it is easier for the programmer to convert it into an actual program.

Link for the ProblemBetween Two Sets – Hacker Rank Solution

Between Two Sets – Hacker Rank Solution


There will be two arrays of integers. Determine all integers that satisfy the following two conditions:

  1. The elements of the first array are all factors of the integer being considered
  2. The integer being considered is a factor of all elements of the second array

These numbers are referred to as being between the two arrays. Determine how many such numbers exist.


Between Two Sets in Algorithm | HackerRank Programming Solutions | HackerRank Problem Solving Solutions in Java [💯Correct]

Function Description

Complete the getTotalX function in the editor below. It should return the number of integers that are betwen the sets.

getTotalX has the following parameter(s):

  • int a[n]: an array of integers
  • int b[m]: an array of integers


  • int: the number of integers that are between the sets

Input Format

image 49

Sample Input

2 3
2 4
16 32 96

Sample Output



2 and 4 divide evenly into 4, 8, 12 and 16.
4, 8 and 16 divide evenly into 16, 32, 96.

4, 8 and 16 are the only three numbers for which each element of a is a factor and each is a factor of all elements of b.

Between Two Sets – Hacker Rank Solution
import java.util.Scanner;

 * @author Techno-RJ
public class BetweenTwoSets {
	public static void main(String[] args) {
		Scanner sc = new Scanner(;
		int n = sc.nextInt();
		int m = sc.nextInt();
		int gcd = 0;

		int a[] = new int[n];
		int b[] = new int[m];

		for (int i = 0; i < n; i++) {
			a[i] = sc.nextInt();
		int lcm = a[0];
		for (int i = 0; i < m; i++) {
			b[i] = sc.nextInt();
			gcd = findGCD(b[i], gcd);
		for (int i = 0; i < n - 1; i++) {
			lcm = (lcm * a[i + 1]) / findGCD(a[i + 1], lcm);
		int count = 0, t = 0;
		for (int i = 1; i <= gcd && t <= gcd; i++) {
			t = lcm * i;
			if (gcd % (lcm * i) == 0) {

	private static int findGCD(int a, int b) {
		return b == 0 ? a : findGCD(b, a % b);


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