Count of primes below N which can be expressed as the sum of two primes

Given an integer N, the task is to find the count of all the primes below N which can be expressed as the sum of two primes.

Examples:

Input: N = 6
Output: 1
5 is the only such prime below 6.
2 + 3 = 5.



Input: N = 11
Output: 2

Approach: Create an array prime[] where prime[i] will store whether i is prime or not using Sieve of Eratosthenes. Now for every prime from the range [1, N – 1], check whether it can be expressed as the sum of two primes using the approach discussed here.

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
const int MAX = 100005;
bool prime[MAX];
  
// Function for Sieve of Eratosthenes
void SieveOfEratosthenes()
{
  
    memset(prime, true, sizeof(prime));
  
    // false here indicates
    // that it is not prime
    prime[0] = false;
    prime[1] = false;
  
    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,
            // set them to non-prime
            for (int i = p * 2; i <= MAX; i += p)
                prime[i] = false;
        }
    }
}
  
// Function to return the count of primes
// less than or equal to n which can be
// expressed as the sum of two primes
int countPrimes(int n)
{
    SieveOfEratosthenes();
  
    // To store the required count
    int cnt = 0;
  
    for (int i = 2; i < n; i++) {
  
        // If the integer is prime and it
        // can be expressed as the sum of
        // 2 and a prime number
        if (prime[i] && prime[i - 2])
            cnt++;
    }
  
    return cnt;
}
  
// Driver code
int main()
{
    int n = 11;
  
    cout << countPrimes(n);
  
    return 0;
}

chevron_right






                        filter_none









Java














                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















// Java implementation of the approach 


class GFG 


{ 


  


static int MAX = 100005; 


static boolean []prime = new boolean[MAX]; 


  


// Function for Sieve of Eratosthenes 


static void SieveOfEratosthenes() 


{ 


  


    for (int i = 0; i < MAX; i++) 


        prime[i] = true; 


  


    // false here indicates 


    // that it is not prime 


    prime[0] = false; 


    prime[1] = false; 


  


    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, 


            // set them to non-prime 


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


                prime[i] = false; 


        } 


    } 


} 


  


// Function to return the count of primes 


// less than or equal to n which can be 


// expressed as the sum of two primes 


static int countPrimes(int n) 


{ 


    SieveOfEratosthenes(); 


  


    // To store the required count 


    int cnt = 0; 


  


    for (int i = 2; i < n; i++) 


    { 


        // If the integer is prime and it 


        // can be expressed as the sum of 


        // 2 and a prime number 


        if (prime[i] && prime[i - 2]) 


            cnt++; 


    } 


    return cnt; 


} 


  


// Driver code 


public static void main(String[] args) 


{ 


    int n = 11; 


  


    System.out.print(countPrimes(n)); 


} 


} 


  


// This code is contributed by 29AjayKumar 



















                        chevron_right







                        filter_none









Python3














                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















# Python3 implementation of the approach 


MAX = 100005


prime = [True for i in range(MAX)] 


  


# Function for Sieve of Eratosthenes 


def SieveOfEratosthenes(): 


  


    # False here indicates 


    # that it is not prime 


    prime[0] = False


    prime[1] = False


  


    for p in range(MAX): 


  


        if(p * p > MAX): 


            break


  


        # If prime[p] is not changed, 


        # then it is a prime 


        if (prime[p]): 


  


            # Update all multiples of p, 


            # set them to non-prime 


            for i in range(2 * p, MAX, p): 


                prime[i] = False


  


# Function to return the count of primes 


# less than or equal to n which can be 


# expressed as the sum of two primes 


def countPrimes(n): 


    SieveOfEratosthenes() 


  


    # To store the required count 


    cnt = 0


  


    for i in range(2, n): 


  


        # If the integer is prime and it 


        # can be expressed as the sum of 


        # 2 and a prime number 


        if (prime[i] and prime[i - 2]): 


            cnt += 1


  


    return cnt 


  


# Driver code 


n = 11


  


print(countPrimes(n)) 


  


# This code is contributed by Mohit Kumar 



















                        chevron_right







                        filter_none









C#














                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















    // C# implementation of the approach 


using System; 


  


class GFG 


{ 


static int MAX = 100005; 


static bool []prime = new bool[MAX]; 


  


// Function for Sieve of Eratosthenes 


static void SieveOfEratosthenes() 


{ 


    for (int i = 0; i < MAX; i++) 


        prime[i] = true; 


  


    // false here indicates 


    // that it is not prime 


    prime[0] = false; 


    prime[1] = false; 


  


    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, 


            // set them to non-prime 


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


                prime[i] = false; 


        } 


    } 


} 


  


// Function to return the count of primes 


// less than or equal to n which can be 


// expressed as the sum of two primes 


static int countPrimes(int n) 


{ 


    SieveOfEratosthenes(); 


  


    // To store the required count 


    int cnt = 0; 


  


    for (int i = 2; i < n; i++) 


    { 


        // If the integer is prime and it 


        // can be expressed as the sum of 


        // 2 and a prime number 


        if (prime[i] && prime[i - 2]) 


            cnt++; 


    } 


    return cnt; 


} 


  


// Driver code 


public static void Main(String[] args) 


{ 


    int n = 11; 


  


    Console.Write(countPrimes(n)); 


} 


} 


  


// This code is contributed by Rajput-Ji 



















                        chevron_right







                        filter_none











Output:


2







GeeksforGeeks has prepared a complete interview preparation course with premium videos, theory, practice problems, TA support and many more features. Please refer Placement 100 for details 





    

        My Personal Notes
        arrow_drop_up
    

    

        
        

            
            
        

    


Recommended Posts:
KunalGupta
Check out this Author's contributed articles.
If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please Improve this article if you find anything incorrect by clicking on  the "Improve Article" button below.