Maximum Primes whose sum is equal to given N

Given a positive integer N > 1. Find the maximum count of prime numbers whose sum is equal to the given N.
Examples: 

Input : N = 5 
Output : 2 
Explanation : 2 and 3 are two prime numbers whose sum is 5.
Input : N = 6 
Output :3 
Explanation : 2, 2, 2 are three prime numbers whose sum is 6.

For the maximum number of primes whose sum is equal to given n, prime numbers must be as small as possible. So, 2 is the smallest possible prime number and is an even number. The next prime number is greater than 2 in 3 which is odd. So, for any given n there are two conditions, either n will be odd or even. 

• Case 1: n is even, In this case, n/2 will be the answer (n/2 number of 2 will result in the sum of n). 
   
• Case 2: n is odd, In this case, floor(n/2) will be the answer ((n-3)/2 number of 2, and one 3 will result in the sum of n). 

Below is the implementation of the above approach: 

// C++ program for above approach

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

// Function to find max count of primes
int maxPrimes(int n)
{
    // if n is even n/2 is required answer
    // if n is odd floor(n/2)  = (int)(n/2) is required answer
    return n / 2;
}

// Driver Code
int main()
{
    int n = 17;

    cout << maxPrimes(n);

    return 0;
}

// Java program for above approach
class GFG
{

// Function to find max count of primes
static int maxPrimes(int n)
{
    // if n is even n/2 is required answer
    // if n is odd floor(n/2) = (int)(n/2)
    // is required answer
    return n / 2;
}

// Driver Code
public static void main(String[] args)
{
    int n = 17;

    System.out.println(maxPrimes(n));
}
}

// This code is contributed
// by Code_Mech

# Python3 program for above approach

# Function to find max count of primes
def maxPrmimes(n):

    # if n is even n/2 is required answer
    # if n is odd floor(n/2) = (int)(n/2) 
    # is required answer
    return n // 2

# Driver code
n = 17
print(maxPrmimes(n))

# This code is contributed
# by Shrikant13

// C# program for above approach
using System;

class GFG
{

// Function to find max count of primes
static int maxPrimes(int n)
{
    // if n is even n/2 is required answer
    // if n is odd floor(n/2) = (int)(n/2)
    // is required answer
    return n / 2;
}

// Driver Code
public static void Main()
{
    int n = 17;

    Console.WriteLine(maxPrimes(n));
}
}

// This code is contributed
// by Akanksha Rai

<?php
// PHP program for above approach

// Function to find max count of primes
function maxPrimes($n)
{
    // if n is even n/2 is required answer
    // if n is odd floor(n/2) = (int)(n/2) is required answer
    return (int)($n / 2);
}

    // Driver Code
    $n = 17;
    echo maxPrimes($n);

// This code is contributed by mits
?>

<script>


// Javascript program for above approach

// Function to find max count of primes
function maxPrimes(n)
{
    // if n is even n/2 is required answer
    // if n is odd floor(n/2)  = (int)(n/2) is required answer
    return parseInt(n / 2);
}

// Driver Code
var n = 17;
document.write( maxPrimes(n));  

</script>    
    

8

Time Complexity: O(1)
Auxiliary Space: O(1), since no extra space has been taken.