user5956891 user5956891 - 16 days ago 9
Java Question

Count number of non-prime pairs that when multiplied form a given number N

A non-prime pair which forms N is 2 different non-prime numbers where the product of the numbers is N.

For example For N = 24 there are 2 good pairs (non-prime pairs that form N) (4,6), (1,24), but (2,12), (3,8) are not good.

Note: for any 2 numbers a and b pair(a,b) = pair(b,a).

There is another condition which states that if the number is a special number, so output = -1 otherwise count the number of non-primes.

Number is called special number if it can be represented as a product of three prime numbers. Example: 12 is a special number because 12=2*2*3;

I tried brute-force approach using https://en.wikipedia.org/wiki/Sieve_of_Eratosthenes ,
which takes O(N*log(log(N)).

"Is there any more optimized way to solve it except brute-force?"

Any idea will be appreciated.

Thanks in advance.

Answer

First of all, Eratosthenes' sieve is O(N*log(log(N)) to list all primes below or equal N (when well implemented).

Second: suppose you factored your number in Q primes with multiplicity which, without sieving, is a process of O(sqrt(N)) at worst (worst: your number is prime). So you have a map of:

  • p0 -> multiplicity m0
  • p1 -> multiplicity m1
  • ...
  • pQ -> multiplicity mQ

How many divisors made from multiplying at least 2 prime factors?

Well, there will be (m0+1)*(m1+1)*...*(mQ+1) - Q of them.

See here for the first product ** ; as for the why - Q - that's the number of distinct divisors that are simply primes - i.e. not composite.

I trust you will be able to code it or... do you need help with that too?


** if you cannot stomach the math notations in the linked, here's a high-school level way to explain this:

  • {1, p0, p02, ...p0m0} are m0+1 ways of generating divisors with the powers of p0
  • {1, p1, p12, ...p1m1} are m1+1 ways of generating divisors with the powers of p1
  • ...
  • {1, p1, p1Q, ...p1mQ} are mQ+1 ways of generating divisors with the powers of pQ

The number of all combinations (as 1 is already included in each set) will be the cardinality of the cartesian product of all the above subsets - thus the product of the individual cardinalities.

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