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

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 : 1
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 : 2

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 `

Output:

```0
```

Time Complexity : O(log(min(n,m)))

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