# 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 ``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;``    ``prime = 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 = 0``    ``dp = 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 = 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 = 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 = ``false``;``    ``prime = ``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 = 0``    ``dp = 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

 ``
Output:
`1`

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

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