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.
Input : N = 5
Output : 2
Explanation : 2 and 3 are two prime numbers whose sum is 5.
Input : N = 6
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:
Time Complexity: O(1)
Auxiliary Space: O(1), since no extra space has been taken.