Find numbers which are multiples of first array and factors of second array

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Given two arrays A[] and B[], the task is to find the integers which are divisible by all the elements of array A[] and divide all the elements of array B[].

Examples:

Input: A[] = {1, 2, 2, 4}, B[] = {16, 32, 64}
Output: 4 8 16
4, 8 and 16 are the only numbers that
are multiples of all the elements of array A[]
and divide all the elements of array B[]
Input: A[] = {2, 3, 6}, B[] = {42, 84}
Output: 6 42

Approach: If X is a multiple of all the elements of the first array then X must be a multiple of the LCM of all the elements of the first array.
Similarly, If X is a factor of all the elements of the second array then it must be a factor of the GCD of all the elements of the second array and such X will exist only if GCD of the second array is divisible by the LCM of the first array.
If it is divisible then X can be any value from the range [LCM, GCD] which is a multiple of LCM and evenly divides GCD.

Below is the implementation of above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the LCM of two numbers
int lcm(int x, int y)
{
    int temp = (x * y) / __gcd(x, y);
    return temp;
}
 
// Function to print the required numbers
void findNumbers(int a[], int n, int b[], int m)
{
 
    // To store the lcm of array a[] elements
    // and the gcd of array b[] elements
    int lcmA = 1, gcdB = 0;
 
    // Finding LCM of first array
    for (int i = 0; i < n; i++)
        lcmA = lcm(lcmA, a[i]);
 
    // Finding GCD of second array
    for (int i = 0; i < m; i++)
        gcdB = __gcd(gcdB, b[i]);
 
    // No such element exists
    if (gcdB % lcmA != 0) {
        cout << "-1";
        return;
    }
 
    // All the multiples of lcmA which are
    // less than or equal to gcdB and evenly
    // divide gcdB will satisfy the conditions
    int num = lcmA;
    while (num <= gcdB) {
        if (gcdB % num == 0)
            cout << num << " ";
        num += lcmA;
    }
}
 
// Driver code
int main()
{
 
    int a[] = { 1, 2, 2, 4 };
    int b[] = { 16, 32, 64 };
 
    int n = sizeof(a) / sizeof(a[0]);
    int m = sizeof(b) / sizeof(b[0]);
 
    findNumbers(a, n, b, m);
 
    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
 
class GFG
{
static int __gcd(int a, int b) 
{ 
    if (b == 0) 
        return a; 
    return __gcd(b, a % b); 
     
}
 
// Function to return the LCM of two numbers
static int lcm(int x, int y)
{
    int temp = (x * y) / __gcd(x, y);
    return temp;
}
 
// Function to print the required numbers
static void findNumbers(int a[], int n, 
                        int b[], int m)
{
 
    // To store the lcm of array a[] elements
    // and the gcd of array b[] elements
    int lcmA = 1, gcdB = 0;
 
    // Finding LCM of first array
    for (int i = 0; i < n; i++)
        lcmA = lcm(lcmA, a[i]);
 
    // Finding GCD of second array
    for (int i = 0; i < m; i++)
        gcdB = __gcd(gcdB, b[i]);
 
    // No such element exists
    if (gcdB % lcmA != 0) 
    {
        System.out.print("-1");
        return;
    }
 
    // All the multiples of lcmA which are
    // less than or equal to gcdB and evenly
    // divide gcdB will satisfy the conditions
    int num = lcmA;
    while (num <= gcdB) 
    {
        if (gcdB % num == 0)
            System.out.print(num + " ");
        num += lcmA;
    }
}
 
// Driver code
public static void main(String[] args)
{
    int a[] = { 1, 2, 2, 4 };
    int b[] = { 16, 32, 64 };
 
    int n = a.length;
    int m = b.length;
 
    findNumbers(a, n, b, m);
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 implementation of the approach 
from math import gcd
 
# Function to return the LCM of two numbers 
def lcm( x, y) :
     
    temp = (x * y) // gcd(x, y); 
    return temp; 
 
# Function to print the required numbers 
def findNumbers(a, n, b, m) : 
 
    # To store the lcm of array a[] elements 
    # and the gcd of array b[] elements 
    lcmA = 1; __gcdB = 0; 
 
    # Finding LCM of first array 
    for i in range(n) : 
        lcmA = lcm(lcmA, a[i]); 
 
    # Finding GCD of second array 
    for i in range(m) : 
        __gcdB = gcd(__gcdB, b[i]); 
 
    # No such element exists 
    if (__gcdB % lcmA != 0) :
        print("-1"); 
        return; 
 
    # All the multiples of lcmA which are 
    # less than or equal to gcdB and evenly 
    # divide gcdB will satisfy the conditions 
    num = lcmA; 
    while (num <= __gcdB) :
        if (__gcdB % num == 0) :
            print(num, end = " "); 
             
        num += lcmA; 
 
# Driver code 
if __name__ == "__main__" : 
 
    a = [ 1, 2, 2, 4 ];
    b = [ 16, 32, 64 ];
     
    n = len(a);
    m = len(b);
     
    findNumbers(a, n, b, m); 
     
# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;
 
class GFG
{
static int __gcd(int a, int b) 
{ 
    if (b == 0) 
        return a; 
    return __gcd(b, a % b); 
}
 
// Function to return the LCM of two numbers
static int lcm(int x, int y)
{
    int temp = (x * y) / __gcd(x, y);
    return temp;
}
 
// Function to print the required numbers
static void findNumbers(int []a, int n, 
                        int []b, int m)
{
 
    // To store the lcm of array a[] elements
    // and the gcd of array b[] elements
    int lcmA = 1, gcdB = 0;
 
    // Finding LCM of first array
    for (int i = 0; i < n; i++)
        lcmA = lcm(lcmA, a[i]);
 
    // Finding GCD of second array
    for (int i = 0; i < m; i++)
        gcdB = __gcd(gcdB, b[i]);
 
    // No such element exists
    if (gcdB % lcmA != 0) 
    {
        Console.Write("-1");
        return;
    }
 
    // All the multiples of lcmA which are
    // less than or equal to gcdB and evenly
    // divide gcdB will satisfy the conditions
    int num = lcmA;
    while (num <= gcdB) 
    {
        if (gcdB % num == 0)
            Console.Write(num + " ");
        num += lcmA;
    }
}
 
// Driver code
public static void Main(String[] args)
{
    int []a = { 1, 2, 2, 4 };
    int []b = { 16, 32, 64 };
 
    int n = a.Length;
    int m = b.Length;
 
    findNumbers(a, n, b, m);
}
}
 
// This code is contributed by 29AjayKumar

Javascript

<script>
 
// Javascript implementation of the approach
 
// Function to find nth centered
// tridecagonal number
function __gcd(a, b) 
{
    if (b == 0) 
        return a; 
         
    return __gcd(b, a % b); 
     
}
 
// Function to return the LCM of two numbers
function lcm(x, y)
{
    var temp = (x * y) / __gcd(x, y);
    return temp;
}
 
// Function to print the required numbers
function findNumbers(a, n, b, m)
{
     
    // To store the lcm of array a[] elements
    // and the gcd of array b[] elements
    var lcmA = 1, gcdB = 0;
 
    // Finding LCM of first array
    for(var i = 0; i < n; i++)
        lcmA = lcm(lcmA, a[i]);
 
    // Finding GCD of second array
    for(var i = 0; i < m; i++)
        gcdB = __gcd(gcdB, b[i]);
 
    // No such element exists
    if (gcdB % lcmA != 0) 
    {
        document.write("-1");
        return;
    }
 
    // All the multiples of lcmA which are
    // less than or equal to gcdB and evenly
    // divide gcdB will satisfy the conditions
    var num = lcmA;
    while (num <= gcdB) 
    {
        if (gcdB % num == 0)
            document.write(num + " ");
             
        num += lcmA;
    }
}
 
// Driver code
var a = [ 1, 2, 2, 4 ];
var b = [ 16, 32, 64 ];
 
var n = a.length;
var m = b.length;
 
findNumbers(a, n, b, m);
 
// This code is contributed by Ankita saini
 
</script>

Output:

4 8 16

Time Complexity: O(max(n,m) * log(min(a, b)))

Auxiliary Space: O(1)

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