Count prime numbers up to N that can be represented as a sum of two prime numbers

Given a positive integer N, the task is to find the count of prime numbers less than or equal to N that can be represented as a sum of two prime numbers.

Examples:

Input: N = 6
Output: 1
Explanation:
5 is the only prime number over the range [1, 6] that can be represented as sum of 2 prime numbers i.e., 5 = 2 + 3, where 2, 3 and 5 are all primes.
Therefore, the count is 2.

Input: N = 14
Output: 3

Naive Approach: The simplest approach to solve the given problem is to consider all possible pairs (i, j) over the range [1, N] and if i and j are prime numbers and (i + j) lies in the range [1, N] then increment the count of prime numbers. After checking for all possible pairs, print the value of the total count obtained. 

Time Complexity: O(N³)
Auxiliary Space: O(1)

Efficient Approach: The above approach can also be optimized based on the following observation:

• Apart from 2, all the prime numbers are odd
• It is not possible to represent a prime number(which is odd) to be represented as a sum of two odd prime numbers, so one of the two prime numbers should be 2.
• So for a prime number X to be a sum of two prime numbers, X – 2 must also be prime.

Follow the steps below to solve the problem:

• Initialize an array, say prime[] of size 10⁵, and populate all the prime numbers till 10⁵ using the Sieve Of Eratosthenes.
• Initialize an auxiliary array dp[] of the size (N + 1) where dp[i] is the count of prime numbers less than or equal to i that can be represented as a sum of 2 prime numbers.
• Iterate over the range [2, N] using the variable i and perform the following steps:
  • Update the value of dp[i – 1] as the sum of dp[i – 1] and dp[i].
  • Check if prime[i] and prime[i – 2] is true, then increment the value of dp[i] by 1.
• After completing the above steps, print the value of dp[N] as the resultant count.

Below is the implementation of the above approach:

C++

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to store all prime numbers
// up to N using Sieve of Eratosthenes
void SieveOfEratosthenes(
    int n, bool prime[])
{
    // Set 0 and 1 as non-prime
    prime[0] = 0;
    prime[1] = 0;

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

        // If p is prime
        if (prime[p] == true) {

            // Set all multiples
            // of p as non-prime
            for (int i = p * p;
                 i <= n; i += p) {
                prime[i] = false;
            }
        }
    }
}

// Function to count prime numbers
// up to N that can be represented
// as the sum of two prime numbers
void countPrime(int n)
{
    // Stores all the prime numbers
    bool prime[n + 1];
    memset(prime, true, sizeof(prime));

    // Update the prime array
    SieveOfEratosthenes(n, prime);

    // Create a dp array of size n + 1
    int dp[n + 1];

    memset(dp, 0, sizeof(dp));

    // Update dp[1] = 0
    dp[1] = 0;

    // Iterate over the range [2, N]
    for (int i = 2; i <= n; i++) {

        // Add the previous count value
        dp[i] += dp[i - 1];

        // Increment dp[i] by 1 if i
        // and (i - 2) are both prime
        if (prime[i] == 1
            && prime[i - 2] == 1) {
            dp[i]++;
        }
    }

    // Print the result
    cout << dp[n];
}

// Driver Code
int main()
{
    int N = 6;
    countPrime(N);

    return 0;
}

Java

// Java approach for the above approach
import java.io.*;
public class GFG{

// Function to store all prime numbers
// up to N using Sieve of Eratosthenes
static void SieveOfEratosthenes(int n, boolean prime[])
{

    // Set 0 and 1 as non-prime
    prime[0] = false;
    prime[1] = false;

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

        // If p is prime
        if (prime[p] == true)
        {

            // Set all multiples
            // of p as non-prime
            for(int i = p * p; i <= n; i += p)
            {
                prime[i] = false;
            }
        }
    }
}

// Function to count prime numbers
// up to N that can be represented
// as the sum of two prime numbers
static void countPrime(int n)
{

    // Stores all the prime numbers
    boolean prime[] = new boolean[n + 1];

    for(int i = 0; i < prime.length; i++)
    {
        prime[i] = true;
    }

    // Update the prime array
    SieveOfEratosthenes(n, prime);

    // Create a dp array of size n + 1
    int dp[] = new int[n + 1];
    for(int i = 0; i < dp.length; i++)
    {
        dp[i] = 0;
    }

    // Update dp[1] = 0
    dp[1] = 0;

    // Iterate over the range [2, N]
    for(int i = 2; i <= n; i++)
    {

        // Add the previous count value
        dp[i] += dp[i - 1];

        // Increment dp[i] by 1 if i
        // and (i - 2) are both prime
        if (prime[i] == true &&
            prime[i - 2] == true)
        {
            dp[i]++;
        }
    }

    // Print the result
    System.out.print(dp[n]);
}

// Driver Code
public static void main(String[] args)
{
    int N = 6;

    countPrime(N);
}
}

// This code is contributed by abhinavjain194

Python3

# Python3 program for the above approach

# Function to store all prime numbers
# up to N using Sieve of Eratosthenes
def SieveOfEratosthenes(n, prime):

    # Set 0 and 1 as non-prime
    prime[0] = 0
    prime[1] = 0

    p = 2

    while p * p <= n:

        # If p is prime
        if (prime[p] == True):

            # Set all multiples
            # of p as non-prime
            for i in range(p * p, n + 1, p):
                prime[i] = False

        p += 1

# Function to count prime numbers
# up to N that can be represented
# as the sum of two prime numbers
def countPrime(n):

    # Stores all the prime numbers
    prime = [True] * (n + 1)

    # Update the prime array
    SieveOfEratosthenes(n, prime)

    # Create a dp array of size n + 1
    dp = [0] * (n + 1)

    # Update dp[1] = 0
    dp[1] = 0

    # Iterate over the range [2, N]
    for i in range(2, n + 1):

        # Add the previous count value
        dp[i] += dp[i - 1]

        # Increment dp[i] by 1 if i
        # and (i - 2) are both prime
        if (prime[i] == 1 and prime[i - 2] == 1):
            dp[i] += 1

    # Print the result
    print(dp[n])

# Driver Code
if __name__ == "__main__":

    N = 6

    countPrime(N)

# This code is contributed by mohit ukasp

C#

// C# program for the above approach
using System;

class GFG{

// Function to store all prime numbers
// up to N using Sieve of Eratosthenes
static void SieveOfEratosthenes(int n, bool[] prime)
{

    // Set 0 and 1 as non-prime
    prime[0] = false;
    prime[1] = false;

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

        // If p is prime
        if (prime[p] == true)
        {

            // Set all multiples
            // of p as non-prime
            for(int i = p * p; i <= n; i += p)
            {
                prime[i] = false;
            }
        }
    }
}

// Function to count prime numbers
// up to N that can be represented
// as the sum of two prime numbers
static void countPrime(int n)
{

    // Stores all the prime numbers
    bool[] prime = new bool[n + 1];

    for(int i = 0; i < prime.Length; i++)
    {
        prime[i] = true;
    }

    // Update the prime array
    SieveOfEratosthenes(n, prime);

    // Create a dp array of size n + 1
    int[] dp = new int[n + 1];
    for(int i = 0; i < dp.Length; i++)
    {
        dp[i] = 0;
    }

    // Update dp[1] = 0
    dp[1] = 0;

    // Iterate over the range [2, N]
    for(int i = 2; i <= n; i++)
    {

        // Add the previous count value
        dp[i] += dp[i - 1];

        // Increment dp[i] by 1 if i
        // and (i - 2) are both prime
        if (prime[i] == true &&
            prime[i - 2] == true)
        {
            dp[i]++;
        }
    }

    // Print the result
    Console.Write(dp[n]);
}

// Driver code
static void Main()
{
    int N = 6;

    countPrime(N);
}
}

// This code is contributed by abhinavjain194

Javascript

<script>

// Javascript program for the above approach

// Function to store all prime numbers
// up to N using Sieve of Eratosthenes
function SieveOfEratosthenes(n, prime)
{

    // Set 0 and 1 as non-prime
    prime[0] = 0;
    prime[1] = 0;

    for(var p = 2; p * p <= n; p++)
    {

        // If p is prime
        if (prime[p] == Boolean(true))
        {

            // Set all multiples
            // of p as non-prime
            for(var i = p * p;i <= n; i += p)
            {
                prime[i] = Boolean(false);
            }
        }
    }
}

// Function to count prime numbers
// up to N that can be represented
// as the sum of two prime numbers
function countPrime(n)
{

    // Stores all the prime numbers
    var prime = new Array(n + 1);
    var x = new Boolean(true);
    prime.fill(x);

    // Update the prime array
    SieveOfEratosthenes(n, prime);

    // Create a dp array of size n + 1
    var dp = new Array(n + 1);
    dp.fill(0);

    // Update dp[1] = 0
    dp[1] = 0;

    // Iterate over the range [2, N]
    for(var i = 2; i <= n; i++)
    {

        // Add the previous count value
        dp[i] += dp[i - 1];

        // Increment dp[i] by 1 if i
        // and (i - 2) are both prime
        if (prime[i] == Boolean(true) &&
            prime[i - 2] == Boolean(true))
        {
            dp[i]++;

        }
    }

    // Print the result
    document.write(dp[n]);
}

// Driver code
var n = 6;

countPrime(n);

// This code is contributed by SoumikMondal

</script>

Output

1

Time Complexity: O(N * log(log(N)))
Auxiliary Space: O(N)

Approach:

Here is the code of above algorithm:

Python3

# Function to check if a number is prime
def isPrime(n):
    if n <= 1:
        return False
    for i in range(2, int(n**0.5) + 1):
        if n % i == 0:
            return False
    return True

# Function to count prime numbers
# up to N that can be represented
# as the sum of two prime numbers
def countPrime(n):
    primes = set()
    count = 0
    for i in range(2, n+1):
        if isPrime(i):
            primes.add(i)
            if (i-2) in primes:
                count += 1
    print(count)

# Driver Code
if __name__ == "__main__":
    N = 6
    countPrime(N)

Output

1

Time Complexity: O(n^(3/2))
Auxiliary Space: O(sqrt(n))