Interview Question

Country: United States

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You only need to know P and K. The list of N integers is irrelevant to the problem.
The first part is to factor P into primes. Then, if all the elements in the resultant list of prime factors are unique, there is a simple combinatorical solution: 1+sum_i=1..K-1 (C(i, L-1)), where C(k,n) = n!/k!(n-k)! and L is length of the list of primes.

I find the general case hard though. When we have duplicates in the list of primes (think of [2, 2, 2, 2, 2] where [2,2],[2,2,2] and [2,2,2],[2,2] result in the same terms, 4*8 = 8*4) I can't think of any better way than to iterate over all 2-way, 3-way, .. K-way splits of the list of prime factors, compute the resultant product terms, and count ignoring repetitions.

- adr July 23, 2018 | Flag Reply

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