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++ 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 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 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# 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 |
<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> |
0
Time Complexity : O(log(min(n,m)))
Auxiliary Space: O(log(min(n, m)))