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Product of every K’th prime number in an array

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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++




// 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;
}


Java




// 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


Python 3




# 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


C#




// 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


Javascript




<script>
// Javascript implementation of the approach
 
let MAX = 1000000;
let prime = new Array(MAX + 1);
function 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.
    prime.fill(true)
 
    // 0 and 1 are not prime numbers
    prime[1] = false;
    prime[0] = false;
 
    for (let 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 (let i = p * 2; i <= MAX; i += p)
                prime[i] = false;
        }
    }
}
 
// compute the answer
function productOfKthPrimes(arr, n, k) {
    // count of primes
    let c = 0;
 
    // product of the primes
    let product = 1;
 
    // traverse the array
    for (let 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;
            }
        }
    }
    document.write(product + "<br>");
}
 
// Driver code
 
// create the sieve
SieveOfEratosthenes();
 
let n = 5, k = 2;
 
let arr = [2, 3, 5, 7, 11];
 
productOfKthPrimes(arr, n, k);
 
// This code is contributed by gfgking.
</script>


Output

21

Complexity Analysis:

  • Time Complexity : O(n)
  • Auxiliary Space: O(MAX)


Last Updated : 07 Sep, 2022
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