Print all the possible prime factors for all the numbers till the given limit

Let’s understand the expected output.

1
2
3
2 <- for 4, since 4 = 2 x 2
5
2 3 <- for 6, since 6 = 2 x 3
7
2 <- for 8, since 8 = 2 x 4
3 <- for 9, since 9 = 3 x 3
2 5 <- for 10, since 10 = 2 x 5

How do we solve this?

Let’s try to approach it like this:

1. Maintain a list of primes we have encountered through out the loop, say primes.
2. For every element in the series till the limit n, say x
3. maintain a flag isPrime to check if there was a factor or not, set to true
4. Loop from 2 till the number x, with a counter i
5. Check if there is any i that is a factor of x
6. if there is this i in the primes, print the number i
7. unset the flag isPrime, since we know that i is a factor of x
8. End of the loop
9. if the counter i is equal to x and isPrime is still set to true, add the current number x to the set of primes we’re maintaining.
10. The same process is repeated for all the numbers till the limit n

Code in C#:

    public class PrimeFactorBuzz
    {
        HashSet<int> primes = new HashSet<int>();

        public void DoPrimeFactors(int n)
        {
            for (int i = 1; i <= n; i++)
            {
                PrimeFactor(i);
            }
        }

        public void PrimeFactor(int number)
        {
            bool isPrime = true;
            int up = number;

            int i = 2;
            while (i < up)
            {
                if (number % i == 0)
                {
                    if (primes.Contains(i))
                        Console.Write($"{i} ");
                    isPrime = false;
                }
                i++;
            }

            if (number == 1)
                Console.Write($"{number} ");
            if (i == up && isPrime)
            {
                Console.Write($"{number} ");
                primes.Add(number);
            }
        }
    }