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++

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// 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;
}

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Java

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// 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);
      
    }
  
}

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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 [tabby title="C#"]

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// 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);
      
    }
  
}

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Output:

10


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Second year Department of Information Technology Jadavpur University

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Improved By : ihritik, rituraj_jain