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++ 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 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 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# 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 |
<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> |
4 8 16
Time Complexity: O(max(n,m) * log(min(a, b)))
Auxiliary Space: O(1)