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Count prime numbers up to N that can be represented as a sum of two prime numbers
  • Last Updated : 28 Apr, 2021

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
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)

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