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 Kth 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 notbool prime[MAX];// Function for Sieve of Eratosthenesvoid 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 arrayint 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 codeint main(){ int k = 3; cout << sumPrime(k); return 0;} |
Java
// Java implementation of the approachimport java.util.*;class GFG {static int MAX = 1000000;// To store whether a number is prime or notstatic boolean []prime = new boolean[MAX];// Function for Sieve of Eratosthenesstatic 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 arraystatic 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 codepublic 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 sqrtMAX = 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#
// C# implementation of the approachusing System;using System.Collections.Generic;class GFG {static int MAX = 1000000;// To store whether a number is prime or notstatic bool []prime = new bool[MAX];// Function for Sieve of Eratosthenesstatic 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 arraystatic 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 codepublic 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 notlet prime = new Array(MAX);// Function for Sieve of Eratosthenesfunction 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 arrayfunction 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:
31
Time Complexity: O(MAX)
Auxiliary Space: O(MAX)
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