Finding GCD of a set of numbers? - c++

Question

So, I was asked this question in an interview. Given a group of numbers (not necessarily distinct), I have to find the multiplication of GCD's of all possible subsets of the given group of numbers.
My approach which I told the interviewer:
1. Recursively generate all possible subsets of the given set.
2a. For a particular subset of the given set:
2b. Find GCD of that subset using the Euclid's Algorithm.
3. Multiply it in the answer being obtained.
Assume GCD of an empty set to be 1.
However, there will be 2^n subsets and this won't work optimally if the n is large. How can I optimise it?

Answer 1

Assume that each array element is an integer in the range 1..U for some U.
Let f(x) be the number of subsets with GCD(x). The solution to the problem is then the sum of d^f(d) for all distinct factors 1 <= d <= U.
Let g(x) be the number of array elements divisible by x.
We have
f(x) = 2^g(x) - SUM(x | y, f(y))
We can compute g(x) in O(n * sqrt(U)) by enumerating all divisors of every array element. f(x) can be computed in O(U log U) from high to low values, by enumerating every multiple of x in the straightforward manner.

Answer 2

Pre - Requisite :
Fermat's little theorem (there is a generalised theorem too) , simple maths , Modular exponentiation
Explanation : Notations : A[] stands for our input array
Clearly the constraints 1<=N<=10^5 , tell me that either you need a O(N * LOG N ) solution , dont try to think DP as its complexity according to me will be N * max(A[i]) i.e. approx. 10^5 * 10 ^ 6 . Why? because you need the GCD of the subsets to make a transition.
Ok , moving on
We can think of clubbing the subsets with the same GCD so as to make the complexity.
So , lets decrement an iterator i from 10^6 to 1 trying to make the set with GCD i !
Now to make the subset with GCD(i) I can club it with any i*j where j is a non negative Integer. Why ?
GCD(i , i*j ) = i
Now ,
We can build a frequency table for any element as the number is quite reachable!
Now , during the contest what I did was , keep the number of subsets with gcd(i) at f2[i]
hence what we do is sum frequency of all elements from j*i where j varies from 1 to floor(i/j)
now the subsets with a common divisor(and not GCD) as i is (2^sum - 1) .
Now we have to subtract from this sum the subsets with GCD greater than i and having i as a common divisor of gcd as i.
This can also be done within the same loop by taking summation of f2[i*j] where j varies from 1 to floor(i/j)
Now the subsets with GCD i equal to 2^sum -1 - summation of f2[ij] Just multiply i ( No . of subsets with GCD i times ) i.e. power ( i , 2^sum -1 - summation of f2[ij] ) . But now to calculate this the exponent part can overflow but you can take its % with given MOD-1 as MOD was prime! (Fermat little theorem) using modular exponentiation
Here is a snippet of my code as I am unsure that can we post the code now!
for(i=max_ele; i >= 1;--i)
{
to_add=F[i];
to_subtract = 0 ;
for(j=2 ;j*i <= max_ele;++j)
{
to_add+=F[j*i];
to_subtract+=F2[j*i];
to_subtract>=(MOD-1)?(to_subtract%=(MOD-1)):0;
}
subsets = (((power(2 , to_add , MOD-1) ) - 1) - to_subtract)%(MOD-1) ;
if(subsets<0)
subsets = (subsets%(MOD-1) +MOD-1)%(MOD-1);
ans = ans * power(i , subsets , MOD);
F2[i]= subsets;
ans %=MOD;
}
I feel like I had complicated the things by using F2, I feel like we can do it without F2 by not taking j = 1. but it's okay I haven't thought about it and this is how I managed to get AC .