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

0 0 3 minutes read

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:

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

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

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!