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 

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)))

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