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  1. Prepare
  2. Algorithms
  3. Implementation
  4. Between Two Sets
  5. Discussions

Between Two Sets

  • yash3shah
     + 2 comments

    EDIT - 1:

    I've updated my code for better readability: I guees this will explain it a lot better.

    import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {
    
    private boolean canADivideI(int i, int[] a, int n){
        boolean answer = false;
        for(int j=0;j<n;j++){
                if(i%a[j] == 0){
                    if(j==n-1){
                        answer = true;
                    }
                }
                else{j=n;}
        }
        return answer;
    }
    
    private boolean canIDivideB(int i, int[] b, int m){
        boolean answer = false;
        for(int k=0;k<m;k++){
            if(b[k]%i ==0){
                if(k==m-1){
                    answer = true;
                }
            }
            else {k=m;}
        }   
        return answer;
    }

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        Solution s = new Solution();
        int n = in.nextInt();
        int m = in.nextInt();
        int[] a = new int[n];
        int max = 100, min = 0, elementsFound = 0;
        for(int a_i=0; a_i < n; a_i++){
            a[a_i] = in.nextInt();
            //Look for Max in A and set that as the MIN 
            if(min < a[a_i]) {min = a[a_i];}
        }
        int[] b = new int[m];
        for(int b_i=0; b_i < m; b_i++){
            b[b_i] = in.nextInt();
            //Look for Min in B and set that as the MAX
            if(max > b[b_i]) {max = b[b_i];}
        }
        for(int i = min; i <=max; i++){
            if(s.canADivideI(i, a, n) && s.canIDivideB(i, b, m)){
                elementsFound++;   
            }
        }
        System.out.println(elementsFound);
    }
}

    I did something simmilar, I guess it is an optimized version.

    import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int m = in.nextInt();
        int[] a = new int[n];
        int max = 100, min = 0, answr = 0;
        for(int a_i=0; a_i < n; a_i++){
            a[a_i] = in.nextInt();
            //Look for Max in A and set that as the MIN 
            if(min < a[a_i]) {min = a[a_i];}
        }
        int[] b = new int[m];
        for(int b_i=0; b_i < m; b_i++){
            b[b_i] = in.nextInt();
            //Look for Min in B and set that as the MAX
            if(max > b[b_i]) {max = b[b_i];}
        }
        for(int i = min; i <= max; i++){
            for(int j = 0; j < n; j++){
                if(i%a[j] == 0){
                    if(j == n-1){
                        for(int k = 0; k < m; k++){
                            if(b[k]%i == 0){
                                if(k == m-1){
                                    answr++;
                                }
                            }
                            else {k = m;}
                        }
                    }
                }
                else{j = n;}
            }
        }
        System.out.println(answr);
    }
}