# 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `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 `

Output:

```21
```

Time Complexity : O(n)

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