# Sum of every K’th prime number in an array

Given an array of integers (less than 10^6), the task is to find the sum of all the prime numbers which appear after every (k-1) prime numbers
i.e. every K’th prime number in the array.

Examples:

```Input : Array : 2, 3, 5, 7, 11 ; n=5; k=2
Output : Sum = 10
Explanation: All the elements of the array are prime. So,
the prime numbers after every K intervals are 3, 7 and their sum is 10.

Input : Array : 41, 23, 12, 17, 18, 19 ; n=6; k=2
Output : Sum = 42
```

A simple approach
We have to traverse the array and find the prime numbers after every (k-1) prime numbers. 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
We will 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 sum.

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 = ``false``; ` `    ``prime = ``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` `solve(``int` `arr[], ``int` `n, ``int` `k) ` `{ ` `    ``// count of primes ` `    ``int` `c = 0; ` ` `  `    ``// sum of the primes ` `    ``long` `long` `int` `sum = 0; ` ` `  `    ``// 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) { ` `                ``sum += arr[i]; ` `                ``c = 0; ` `            ``} ` `        ``} ` `    ``} ` `    ``cout << sum << endl; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` ` `  `    ``// create the sieve ` `    ``SieveOfEratosthenes(); ` ` `  `    ``int` `n = 5, k = 2; ` ` `  `    ``int` `arr[n] = { 2, 3, 5, 7, 11 }; ` ` `  `    ``solve(arr, n, k); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` ` `  `class` `GFG ` `{ ` `       `  `    ``static` `final` `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. ` `        ``for``(``int` `i=``0``;i<=MAX;i++) ` `            ``prime[i]=``true``; ` `     `  `        ``// 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 ` `    ``static` `void` `solve(``int` `[]arr, ``int` `n, ``int` `k) ` `    ``{ ` `        ``// count of primes ` `        ``int` `c = ``0``; ` `     `  `        ``// sum of the primes ` `        ``long`  `sum = ``0``; ` `     `  `        ``// 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``) { ` `                    ``sum += arr[i]; ` `                    ``c = ``0``; ` `                ``} ` `            ``} ` `        ``} ` `        ``System.out.println(sum); ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void` `main(String []args) ` `    ``{ ` `     `  `        ``// create the sieve ` `        ``SieveOfEratosthenes(); ` `     `  `        ``int` `n = ``5``, k = ``2``; ` `     `  `        ``int` `[]arr = { ``2``, ``3``, ``5``, ``7``, ``11` `}; ` `     `  `        ``solve(arr, n, k); ` `     `  `    ``} ` ` `  `} `

## Python3

 `# Python3 implementation of the approach  ` `def` `SieveOfEratosthenes():  ` ` `  `    ``# 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` `solve(arr, n, k):  ` ` `  `    ``# count of primes  ` `    ``c ``=` `0` ` `  `    ``# sum of the primes  ` `    ``Sum` `=` `0` ` `  `    ``# Traverse the array  ` `    ``for` `i ``in` `range``(``0``, 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``:  ` `                ``Sum` `+``=` `arr[i]  ` `                ``c ``=` `0` `             `  `    ``print``(``Sum``) ` ` `  `# Driver code  ` `if` `__name__ ``=``=` `"__main__"``: ` ` `  `    ``MAX` `=` `1000000` `    ``prime ``=` `[``True``] ``*` `(``MAX` `+` `1``)  ` ` `  `    ``# Create the sieve  ` `    ``SieveOfEratosthenes()  ` ` `  `    ``n, k ``=` `5``, ``2` `    ``arr ``=` `[``2``, ``3``, ``5``, ``7``, ``11``]  ` ` `  `    ``solve(arr, n, k)  ` `     `  `# This code is contributed by Rituraj Jain `

## C#

 `// C# implementation of the approach ` ` `  `using` `System; ` `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. ` `        ``for``(``int` `i=0;i<=MAX;i++) ` `            ``prime[i]=``true``; ` `     `  `        ``// 0 and 1 are not prime numbers ` `        ``prime = ``false``; ` `        ``prime = ``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 ` `    ``static` `void` `solve(``int` `[]arr, ``int` `n, ``int` `k) ` `    ``{ ` `        ``// count of primes ` `        ``int` `c = 0; ` `     `  `        ``// sum of the primes ` `        ``long`  `sum = 0; ` `     `  `        ``// 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) { ` `                    ``sum += arr[i]; ` `                    ``c = 0; ` `                ``} ` `            ``} ` `        ``} ` `        ``Console.WriteLine(sum); ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `     `  `        ``// create the sieve ` `        ``SieveOfEratosthenes(); ` `     `  `        ``int` `n = 5, k = 2; ` `     `  `        ``int` `[]arr = { 2, 3, 5, 7, 11 }; ` `     `  `        ``solve(arr, n, k); ` `     `  `    ``} ` ` `  `} `

Output:

```10
``` My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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Improved By : ihritik, rituraj_jain