# 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=9Output :1

First array would be = [ 2, 4, 6, 8 ]

Second array would be = [ 3, 6, 9 ]

6 is the only common elementInput :n=1 m=2 k=5Output :2

**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 <bits/stdc++.h>` `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

`<script>` `// javascript implementation of the above approach` `// Recursive function to find` `// gcd using euclidean algorithm` `function` `gcd(a, b)` `{` ` ` `if` `(a == 0)` ` ` `return` `b;` ` ` `return` `gcd(b % a, a);` `}` `// Function to find lcm` `// of two numbers using gcd` `function` `lcm(n, m)` `{` ` ` `return` `(n * m) / gcd(n, m);` `}` `// Driver code` `var` `n = 2, m = 3, k = 5;` `document.write( parseInt(k / lcm(n, m)));` `// This code is contributed by Amit Katiyar` `</script>` |

**Output:**

0

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