Related Articles

# Count common elements in two arrays containing multiples of N and M

• Last Updated : 15 Mar, 2021

Given two arrays such that the first array contains multiples of an integer n which are less than or equal to k and similarly, the second array contains multiples of an integer m which are less than or equal to k.
The task is to find the number of common elements between the arrays.
Examples:

Input :n=2 m=3 k=9
Output :
First array would be = [ 2, 4, 6, 8 ]
Second array would be = [ 3, 6, 9 ]
6 is the only common element
Input :n=1 m=2 k=5
Output :

Approach :
Find the LCM of n and m .As LCM is the least common multiple of n and m, all the multiples of LCM would be common in both the arrays. The number of multiples of LCM which are less than or equal to k would be equal to k/(LCM(m, n)).
To find the LCM first calculate the GCD of two numbers using the Euclidean algorithm and lcm of n, m is n*m/gcd(n, m).
Below is the implementation of the above approach:

## C++

 `// C++ implementation of the above approach``#include ` `using` `namespace` `std;` `// Recursive function to find``// gcd using euclidean algorithm``int` `gcd(``int` `a, ``int` `b)``{``    ``if` `(a == 0)``        ``return` `b;``    ``return` `gcd(b % a, a);``}` `// Function to find lcm``// of two numbers using gcd``int` `lcm(``int` `n, ``int` `m)``{``    ``return` `(n * m) / gcd(n, m);``}` `// Driver code``int` `main()``{``    ``int` `n = 2, m = 3, k = 5;` `    ``cout << k / lcm(n, m) << endl;` `    ``return` `0;``}`

## Java

 `// Java implementation of the above approach``import` `java.util.*;``import` `java.lang.*;``import` `java.io.*;` `class` `GFG``{` `// Recursive function to find``// gcd using euclidean algorithm``static` `int` `gcd(``int` `a, ``int` `b)``{``    ``if` `(a == ``0``)``        ``return` `b;``    ``return` `gcd(b % a, a);``}` `// Function to find lcm``// of two numbers using gcd``static` `int` `lcm(``int` `n, ``int` `m)``{``    ``return` `(n * m) / gcd(n, m);``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``int` `n = ``2``, m = ``3``, k = ``5``;` `    ``System.out.print( k / lcm(n, m));``}``}` `// This code is contributed by mohit kumar 29`

## Python3

 `# Python3 implementation of the above approach` `# Recursive function to find``# gcd using euclidean algorithm``def` `gcd(a, b) :` `    ``if` `(a ``=``=` `0``) :``        ``return` `b;``        ` `    ``return` `gcd(b ``%` `a, a);` `# Function to find lcm``# of two numbers using gcd``def` `lcm(n, m) :` `    ``return` `(n ``*` `m) ``/``/` `gcd(n, m);`  `# Driver code``if` `__name__ ``=``=` `"__main__"` `:` `    ``n ``=` `2``; m ``=` `3``; k ``=` `5``;` `    ``print``(k ``/``/` `lcm(n, m));` `# This code is contributed by AnkitRai01`

## C#

 `// C# implementation of the above approach``using` `System;``    ` `class` `GFG``{` `// Recursive function to find``// gcd using euclidean algorithm``static` `int` `gcd(``int` `a, ``int` `b)``{``    ``if` `(a == 0)``        ``return` `b;``    ``return` `gcd(b % a, a);``}` `// Function to find lcm``// of two numbers using gcd``static` `int` `lcm(``int` `n, ``int` `m)``{``    ``return` `(n * m) / gcd(n, m);``}` `// Driver code``public` `static` `void` `Main(String[] args)``{``    ``int` `n = 2, m = 3, k = 5;` `    ``Console.WriteLine( k / lcm(n, m));``}``}` `// This code is contributed by Princi Singh`

## Javascript

 ``
Output:
`0`

Time Complexity : O(log(min(n,m))) My Personal Notes arrow_drop_up