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 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#
// 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)