Given an even number ( greater than 2 ), return two prime numbers whose sum will be equal to given number. NOTE A solution will always exist. read Goldbach’s conjecture Example: Input : 4 Output: 2 + 2 = 4 If there are more than one solutions possible, return the lexicographically smaller solution. If [a, b] is one solution with a <= b, and [c,d…

PrimeSum.cpp

bool isPrime(int n)
{
    for(int i=2; i*i<=n; ++i)
    {
        if(n%i==0)
            return false;
    }
    return true;
}

vector<int> Solution::primesum(int A) {
    vector<int> result;
    if(A<3)
        return result;
    vector<bool> arr(A+1, 0); //vector<int> causes judge memory overflow.
    
    for (int i=2; i<=A; ++i)
    {
        if(arr[i]==0)
        {
            //prime.emplace_back(i);
            if(isPrime(A-i))
            {
                result.emplace_back(i);
                result.emplace_back(A-i);
                break;
            }
            for(int j=i; i*j<=A; ++j)
                arr[i*j] = 1;
        }   
    }
    return result;
}

this could be one
`int is_prime(int p) {
int prime = 1;
for(int i = 2; i < p/2; i++){
if(p%i == 0) prime = 0;
}
if(prime) return 1;
else return 0;
}
vector Solution::primesum(int A) {
vector res;
if(A <= 3) return res;

int diff = 0;
for(int i = 2; i < A; i++) {
    if(is_prime(i)) {
        diff = A - i;
        if(is_prime(diff)) {
            res.push_back(i);
            res.push_back(diff);
            break;
        }
    }
}
return res;

}`