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++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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    // 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

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Output:

2

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