Find the sum of prime numbers in the Kth array

Last Updated : 12 Jul, 2025

Given K arrays where the first array contains the first prime number, the second array contains the next 2 primes and the third array contains the next 3 primes and so on. The task is to find the sum of primes in the K^th array.
Examples:

Input: K = 3
Output: 31
arr1[] = {2}
arr[] = {3, 5}
arr[] = {7, 11, 13}
7 + 11 + 13 = 31
Input: K = 2
Output: 8

Approach: Sieve of Eratosthenes can be used to find all the prime upto the required element. And the count of prime numbers in the arrays from 1 to K - 1 will be cnt = 1 + 2 + 3 + ... + (K - 1) = (K * (K - 1)) / 2. Now, starting from the (cnt + 1)^th prime from the sieve array, start adding all the primes until exactly K primes are added then print the 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

// To store whether a number is prime or not
bool prime[MAX];

// Function for Sieve of Eratosthenes
void SieveOfEratosthenes()
{
    // Create a boolean array "prime[0..n]" and initialize
    // all entries it 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;

    for (int p = 2; p * p < MAX; p++) {

        // If prime[p] is not changed then it is a prime
        if (prime[p]) {

            // Update all multiples of p greater than or
            // equal to the square of it
            // numbers which are multiple of p and are
            // less than p^2 are already been marked.
            for (int i = p * p; i < MAX; i += p)
                prime[i] = false;
        }
    }
}

// Function to return the sum of
// primes in the Kth array
int sumPrime(int k)
{

    // Update vector v to store all the
    // prime numbers upto MAX
    SieveOfEratosthenes();
    vector<int> v;
    for (int i = 2; i < MAX; i++) {
        if (prime[i])
            v.push_back(i);
    }

    // To store the sum of primes
    // in the kth array
    int sum = 0;

    // Count of primes which are in
    // the arrays from 1 to k - 1
    int skip = (k * (k - 1)) / 2;

    // k is the number of primes
    // in the kth array
    while (k > 0) {
        sum += v[skip];
        skip++;

        // A prime has been
        // added to the sum
        k--;
    }

    return sum;
}

// Driver code
int main()
{
    int k = 3;

    cout << sumPrime(k);

    return 0;
}

Java

// Java implementation of the approach
import java.util.*;

class GFG 
{

static int MAX = 1000000;

// To store whether a number is prime or not
static boolean []prime = new boolean[MAX];

// Function for Sieve of Eratosthenes
static void SieveOfEratosthenes()
{
    // Create a boolean array "prime[0..n]" and 
    // initialize all entries it 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;

    for (int p = 2; p * p < MAX; p++) 
    {

        // If prime[p] is not changed
        // then it is a prime
        if (prime[p]) 
        {

            // Update all multiples of p greater than or
            // equal to the square of it
            // numbers which are multiple of p and are
            // less than p^2 are already been marked.
            for (int i = p * p; i < MAX; i += p)
                prime[i] = false;
        }
    }
}

// Function to return the sum of
// primes in the Kth array
static int sumPrime(int k)
{

    // Update vector v to store all the
    // prime numbers upto MAX
    SieveOfEratosthenes();
    Vector<Integer> v = new Vector<>();
    for (int i = 2; i < MAX; i++) 
    {
        if (prime[i])
            v.add(i);
    }

    // To store the sum of primes
    // in the kth array
    int sum = 0;

    // Count of primes which are in
    // the arrays from 1 to k - 1
    int skip = (k * (k - 1)) / 2;

    // k is the number of primes
    // in the kth array
    while (k > 0)
    {
        sum += v.get(skip);
        skip++;

        // A prime has been
        // added to the sum
        k--;
    }

    return sum;
}

// Driver code
public static void main(String[] args)
{
    int k = 3;

    System.out.println(sumPrime(k));
}
}

// This code is contributed by Rajput-Ji

Python3

# Python3 implementation of the approach 
from math import sqrt

MAX = 1000000

# Create a boolean array "prime[0..n]" and 
# initialize all entries it as true. 
# A value in prime[i] will finally be false 
# if i is Not a prime, else true. 
prime = [True] * MAX

# Function for Sieve of Eratosthenes 
def SieveOfEratosthenes() :

    for p in range(2, int(sqrt(MAX)) + 1) : 

        # If prime[p] is not changed
        # then it is a prime 
        if (prime[p]) :

            # Update all multiples of p greater than or 
            # equal to the square of it 
            # numbers which are multiple of p and are 
            # less than p^2 are already been marked. 
            for i in range(p * p, MAX, p) :
                prime[i] = False; 

# Function to return the sum of 
# primes in the Kth array 
def sumPrime(k) : 

    # Update vector v to store all the 
    # prime numbers upto MAX 
    SieveOfEratosthenes(); 
    v = []; 
    for i in range(2, MAX) :
        if (prime[i]) :
            v.append(i); 

    # To store the sum of primes 
    # in the kth array 
    sum = 0; 

    # Count of primes which are in 
    # the arrays from 1 to k - 1 
    skip = (k * (k - 1)) // 2; 

    # k is the number of primes 
    # in the kth array 
    while (k > 0) :
        sum += v[skip]; 
        skip += 1; 

        # A prime has been 
        # added to the sum 
        k -= 1; 

    return sum; 

# Driver code 
if __name__ == "__main__" :
    
    k = 3;
    
    print(sumPrime(k)); 

# This code is contributed by AnkitRai01

// C# implementation of the approach
using System;
using System.Collections.Generic;

class GFG 
{
static int MAX = 1000000;

// To store whether a number is prime or not
static bool []prime = new bool[MAX];

// Function for Sieve of Eratosthenes
static void SieveOfEratosthenes()
{
    // Create a boolean array "prime[0..n]" and 
    // initialize all entries it 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;

    for (int p = 2; p * p < MAX; p++) 
    {

        // If prime[p] is not changed
        // then it is a prime
        if (prime[p]) 
        {

            // Update all multiples of p greater than or
            // equal to the square of it
            // numbers which are multiple of p and are
            // less than p^2 are already been marked.
            for (int i = p * p; i < MAX; i += p)
                prime[i] = false;
        }
    }
}

// Function to return the sum of
// primes in the Kth array
static int sumPrime(int k)
{

    // Update vector v to store all the
    // prime numbers upto MAX
    SieveOfEratosthenes();
    List<int> v = new List<int>();
    for (int i = 2; i < MAX; i++) 
    {
        if (prime[i])
            v.Add(i);
    }

    // To store the sum of primes
    // in the kth array
    int sum = 0;

    // Count of primes which are in
    // the arrays from 1 to k - 1
    int skip = (k * (k - 1)) / 2;

    // k is the number of primes
    // in the kth array
    while (k > 0)
    {
        sum += v[skip];
        skip++;

        // A prime has been
        // added to the sum
        k--;
    }

    return sum;
}

// Driver code
public static void Main(String[] args)
{
    int k = 3;

    Console.WriteLine(sumPrime(k));
}
}

// This code is contributed by PrinciRaj1992

JavaScript

<script>
// Javascript implementation of the approach\

const MAX = 1000000;

// To store whether a number is prime or not
let prime = new Array(MAX);

// Function for Sieve of Eratosthenes
function SieveOfEratosthenes()
{
    // Create a boolean array "prime[0..n]" and initialize
    // all entries it as true. A value in prime[i] will
    // finally be false if i is Not a prime, else true.
    for (let i = 0; i < MAX; i++)
        prime[i] = true;

    for (let p = 2; p * p < MAX; p++) {

        // If prime[p] is not changed then it is a prime
        if (prime[p]) {

            // Update all multiples of p greater than or
            // equal to the square of it
            // numbers which are multiple of p and are
            // less than p^2 are already been marked.
            for (let i = p * p; i < MAX; i += p)
                prime[i] = false;
        }
    }
}

// Function to return the sum of
// primes in the Kth array
function sumPrime(k)
{

    // Update vector v to store all the
    // prime numbers upto MAX
    SieveOfEratosthenes();
    let v = [];
    for (let i = 2; i < MAX; i++) {
        if (prime[i])
            v.push(i);
    }

    // To store the sum of primes
    // in the kth array
    let sum = 0;

    // Count of primes which are in
    // the arrays from 1 to k - 1
    let skip = parseInt((k * (k - 1)) / 2);

    // k is the number of primes
    // in the kth array
    while (k > 0) {
        sum += v[skip];
        skip++;

        // A prime has been
        // added to the sum
        k--;
    }

    return sum;
}

// Driver code
    let k = 3;

    document.write(sumPrime(k));

</script>

Output:

Time Complexity: O(MAX)

Auxiliary Space: O(MAX)

Analysis of Algorithms

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