user5956891 - 1 year ago 94
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.

1<=N<=10^6

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.

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) - 2*Q)/2` of them [correction here]

Now, why `-2*Q` and `/2`?

The problem required pairs no matter the symmetry of them. So:

• when I'm excluding one prime divisor, I'm actually excluding 2 pairs - (px * N/px) and (N/px * px). So, when I'm excluding `Q` primes as non-composite factors, I'm excluding `2*Q` pairs
• similarly, when including something in the result but irrespective of the order of terms, I'll have the two (px * N/px) and (N/px * px) pairs counted as 1.

** 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.

Code - Java

``````static void factor(long N, Map<Long, Short> primesWithMultiplicity) {
// some arg checking and trivial cases
if(N<0) N=-N;
if(0==N) {
throw new IllegalArgumentException(
"Are you kidding me? Every number divides 0, "+
"you really want them all listed?"
);
}
if(1==N) {
primesWithMultiplicity.put(1,(short)1);
return;
}

// don't try divisors higher than sqrt(N),
// if they would have been detected by their composite-complement
for(long div=2; div*div < N; ) {
short multiplicity=0;
while((N % div)==0) {
multiplicity++;
N /= div; // reduce N
}
if(multiplicity>0) {
primesWithMultiplicity.put(div, multiplicity);
}
div+= (div == 2 ? 1 : 2); // from 2 to 3, but then going only on odd numbers
}
// done.. well almost, if N is prime, then
// trying to divide up to sqrt(N) will lead an empty result. But,
// in this case, N will still be at original value (as opposed
// to being 1 if complete factored)
if(N>1) {
primesWithMultiplicity.put(N, 1);
}
}

int countCompositeDivisors(long N) {
HashMap<Long, Short> factoringResults=new HashMap<>;
factor(N, factoringResults);
int ret=1;
for(short multiplicity : factoringResults.values()) {
ret*=(1+multiplicity);
}
return (ret-2*factoringResults.size())/2;
}
``````
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