Sum 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 sum of every k’th prime number in the array.

Examples:

Input: arr[] = {2, 3, 5, 7, 11}, k = 2 
Output: 10 
All the elements of the array are prime. So, the prime numbers after every K (i.e. 2) interval are 3, 7 and their sum is 10.

Input: arr[] = {11, 13, 15, 17, 19}, k = 2 
Output: 32 

Simple Approach: Traverse the array and find every K’th prime number in the array and calculate the running sum. 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 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 SumOfKthPrimes(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 arr[] = { 2, 3, 5, 7, 11 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 2;

    SumOfKthPrimes(arr, n, k);

    return 0;
}

Java

// Java implementation of the approach

public 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.
        for (int i = 0; i < prime.length; 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 SumOfKthPrimes(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 arr[] = {2, 3, 5, 7, 11};
        int n = arr.length;
        int k = 2;
        SumOfKthPrimes(arr, n, k);
    }
}

Python3

# Python3 implementation of the approach
MAX = 100000;
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]):

            # Update all multiples of p
            i = p * 2;
            while(i <= MAX):
                prime[i] = False;
                i += p;
        p += 1;

# compute the answer
def SumOfKthPrimes(arr, n, k):

    # count of primes
    c = 0;

    # sum of the primes
    sum = 0;

    # 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):
                sum += arr[i];
                c = 0;

    print(sum);

# Driver code

# create the sieve
SieveOfEratosthenes();

arr = [ 2, 3, 5, 7, 11 ];
n = len(arr);
k = 2;

SumOfKthPrimes(arr, n, k);

# This code is contributed by mits

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] = 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] == false)
        {

            // Update all multiples of p
            for (int i = p * 2;
                     i <= MAX; i += p)
                prime[i] = true;
        }
    }
}

// compute the answer
static void SumOfKthPrimes(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.Console.WriteLine(sum);
}

// Driver code
static void Main()
{
    // create the sieve
    SieveOfEratosthenes();

    int[] arr = new int[] { 2, 3, 5, 7, 11 };
    int n = arr.Length;
    int k = 2;

    SumOfKthPrimes(arr, n, k);
}
}

// This code is contributed by mits

PHP

<?php
// PHP implementation of the approach
$MAX = 100000;
$prime = array_fill(0, $MAX + 1, true);
function SieveOfEratosthenes()
{
    global $MAX, $prime;

    // 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;

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

// compute the answer
function SumOfKthPrimes($arr, $n, $k)
{
    global $MAX, $prime;

    // count of primes
    $c = 0;

    // sum of the primes
    $sum = 0;

    // traverse the array
    for ($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;
            }
        }
    }
    echo $sum."\n";
}

// Driver code

// create the sieve
SieveOfEratosthenes();

$arr = array( 2, 3, 5, 7, 11 );
$n = sizeof($arr);
$k = 2;

SumOfKthPrimes($arr, $n, $k);

// This code is contributed by mits
?>

Javascript

<script>
// Javascript implementation of the approach
let MAX = 100000;
let prime = new Array(MAX + 1).fill(true);
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.

    // 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 SumOfKthPrimes(arr, n, k) {
    // count of primes
    let c = 0;

    // sum of the primes
    let sum = 0;

    // 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) {
                sum += arr[i];
                c = 0;
            }
        }
    }
    document.write(sum + "<br>");
}

// Driver code

// create the sieve
SieveOfEratosthenes();

let arr = new Array(2, 3, 5, 7, 11);
let n = arr.length;
let k = 2;

SumOfKthPrimes(arr, n, k);

// This code is contributed by gfgking
</script>

Output

10

Complexity Analysis:

• Time Complexity: O(n + MAX²)
• Auxiliary Space: O(MAX)