Product of every K’th prime number in an array

Given an integer ‘k’ and an array of integers ‘arr’ (less than 10^6), the task is to find the product of every K’th prime number in the array.

Examples:

Input: arr = {2, 3, 5, 7, 11}, k = 2
Output: 21
All the elements of the array are prime. So, the prime numbers after every K (i.e. 2) interval are 3, 7 and their product is 21.



Input: arr = {41, 23, 12, 17, 18, 19}, k = 2
Output: 437

A simple approach: Traverse the array and find every K’th prime number in the array and calculate the running product. In this way, we’ll have to check every element of the array whether it is prime or not which will take more time as the size of the array increases.

Efficient approach: Create a sieve which will store whether a number is prime or not. Then, it can be used to check a number against prime in O(1) time. In this way, we only have to keep track of every K’th prime number and maintain the running product.

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
#define MAX 1000000
bool prime[MAX + 1];
void SieveOfEratosthenes()
{
    // Create a boolean array "prime[0..n]"
    // and initialize all the entries as true.
    // A value in prime[i] will finally be false
    // if i is Not a prime, else true.
    memset(prime, true, sizeof(prime));
  
    // 0 and 1 are not prime numbers
    prime[1] = false;
    prime[0] = false;
  
    for (int p = 2; p * p <= MAX; p++) {
  
        // If prime[p] is not changed, 
        // then it is a prime
        if (prime[p] == true) {
  
            // Update all multiples of p
            for (int i = p * 2; i <= MAX; i += p)
                prime[i] = false;
        }
    }
}
  
// compute the answer
void productOfKthPrimes(int arr[], int n, int k)
{
    // count of primes
    int c = 0;
  
    // product of the primes
    long long int product = 1;
  
    // traverse the array
    for (int i = 0; i < n; i++) {
  
        // if the number is a prime
        if (prime[arr[i]]) {
  
            // increase the count
            c++;
  
            // if it is the K'th prime
            if (c % k == 0) {
                product *= arr[i];
                c = 0;
            }
        }
    }
    cout << product << endl;
}
  
// Driver code
int main()
{
  
    // create the sieve
    SieveOfEratosthenes();
  
    int n = 5, k = 2;
  
    int arr[n] = { 2, 3, 5, 7, 11 };
  
    productOfKthPrimes(arr, n, k);
  
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java implementation of the approach
  
class GFG
{
static int MAX=1000000;
static boolean[] prime=new boolean[MAX + 1];
static void SieveOfEratosthenes()
{
    // Create a boolean array "prime[0..n]"
    // and initialize all the entries as true.
    // A value in prime[i] will finally be false
    // if i is Not a prime, else true.
    //memset(prime, true, sizeof(prime));
  
    // 0 and 1 are not prime numbers
    prime[1] = true;
    prime[0] = true;
  
    for (int p = 2; p * p <= MAX; p++) {
  
        // If prime[p] is not changed, 
        // then it is a prime
        if (prime[p] == false) {
  
            // Update all multiples of p
            for (int i = p * 2; i <= MAX; i += p)
                prime[i] = true;
        }
    }
}
  
// compute the answer
static void productOfKthPrimes(int arr[], int n, int k)
{
    // count of primes
    int c = 0;
  
    // product of the primes
    int product = 1;
  
    // traverse the array
    for (int i = 0; i < n; i++) {
  
        // if the number is a prime
        if (!prime[arr[i]]) {
  
            // increase the count
            c++;
  
            // if it is the K'th prime
            if (c % k == 0) {
                product *= arr[i];
                c = 0;
            }
        }
    }
    System.out.println(product);
}
  
// Driver code
public static void main(String[] args)
{
  
    // create the sieve
    SieveOfEratosthenes();
  
    int n = 5, k = 2;
   
    int[] arr=new int[]{ 2, 3, 5, 7, 11 };
   
    productOfKthPrimes(arr, n, k);
}
}
// This code is contributed by mits

chevron_right


Python 3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python 3 implementation of the approach
  
MAX = 1000000
prime = [True]*(MAX + 1)
def SieveOfEratosthenes():
      
    # Create a boolean array "prime[0..n]"
    # and initialize all the entries as true.
    # A value in prime[i] will finally be false
    # if i is Not a prime, else true.
      
  
    # 0 and 1 are not prime numbers
    prime[1] = False;
    prime[0] = False;
  
    p = 2
    while p * p <= MAX:
  
        # If prime[p] is not changed, 
        # then it is a prime
        if (prime[p] == True):
  
            # Update all multiples of p
            for i in range(p * 2, MAX+1, p):
                prime[i] = False
        p+=1
  
# compute the answer
def productOfKthPrimes(arr, n, k):
  
    # count of primes
    c = 0
  
    # product of the primes
    product = 1
  
    # traverse the array
    for i in range( n):
  
        # if the number is a prime
        if (prime[arr[i]]):
  
            # increase the count
            c+=1
  
            # if it is the K'th prime
            if (c % k == 0) :
                product *= arr[i]
                c = 0
  
    print(product)
  
# Driver code
if __name__ == "__main__":
  
    # create the sieve
    SieveOfEratosthenes()
  
    n = 5
    k = 2
  
    arr = [ 2, 3, 5, 7, 11 ]
  
    productOfKthPrimes(arr, n, k)
  
# This code is contributed by ChitraNayal

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# implementation of the approach
class GFG
{
static int MAX = 1000000;
static bool[] prime = new bool[MAX + 1];
static void SieveOfEratosthenes()
{
    // Create a boolean array "prime[0..n]"
    // and initialize all the entries as 
    // true. A value in prime[i] will 
    // finally be false if i is Not a prime,
    // else true.
  
    // 0 and 1 are not prime numbers
    prime[1] = true;
    prime[0] = true;
  
    for (int p = 2; p * p <= MAX; p++)
    {
  
        // If prime[p] is not changed, 
        // then it is a prime
        if (prime[p] == false
        {
  
            // Update all multiples of p
            for (int i = p * 2;
                     i <= MAX; i += p)
                prime[i] = true;
        }
    }
}
  
// compute the answer
static void productOfKthPrimes(int[] arr,
                               int n, int k)
{
    // count of primes
    int c = 0;
  
    // product of the primes
    int product = 1;
  
    // traverse the array
    for (int i = 0; i < n; i++) 
    {
  
        // if the number is a prime
        if (!prime[arr[i]]) 
        {
  
            // increase the count
            c++;
  
            // if it is the K'th prime
            if (c % k == 0) 
            {
                product *= arr[i];
                c = 0;
            }
        }
    }
    System.Console.WriteLine(product);
}
  
// Driver code
static void Main()
{
  
    // create the sieve
    SieveOfEratosthenes();
  
    int n = 5, k = 2;
  
    int[] arr=new int[]{ 2, 3, 5, 7, 11 };
  
    productOfKthPrimes(arr, n, k);
}
}
  
// This code is contributed by mits

chevron_right


Output:

21

Time Complexity : O(n)



My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.