Skip to content
Related Articles

Related Articles

Improve Article

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 <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)))
 




My Personal Notes arrow_drop_up
Recommended Articles
Page :