user5956891 - 11 months ago 74

Java Question

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 Source

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:

- p
_{0}-> multiplicity m_{0} - p
_{1}-> multiplicity m_{1} - ...
- p
_{Q}-> multiplicity m_{Q}

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, p0
^{2}, ...p0^{m0}} are`m0+1`

ways of generating divisors with the powers of`p0`

- {1, p1, p1
^{2}, ...p1^{m1}} are`m1+1`

ways of generating divisors with the powers of`p1`

- ...
- {1, p1, p1
^{Q}, ...p1^{mQ}} 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.