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 number 
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 that 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 <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 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[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;
                }
            }
        }
        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);

    }

}

Javascript

<script>

// Javascript program to find next
// greater number than N

MAX = 1000000;
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 (var 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 (var i = p * 2; i <= MAX; i += p)
                prime[i] = false;
        }
    }
}

// compute the answer
function solve(arr, n, k)
{
    // count of primes
    var c = 0;

    // sum of the primes
    var sum = 0;

    // traverse the array
    for (var 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;
            }
        }
    }
    document.write( sum + "<br>")
}

SieveOfEratosthenes();
var n = 5, k = 2;
var arr = [ 2, 3, 5, 7, 11 ];
solve(arr, n, k);

// This code is contributed by SoumikMondal

</script>

Output

10

complexity Analysis:

• Time Complexity: O(n + MAX^3/2)
• Auxiliary Space: O(MAX)