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# 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 :
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)))

Auxiliary Space: O(log(min(n, m)))

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