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Thank you very much, actually i understood the theory, i also made a solution i am not entirely happy with, but i was actually trying to understand your recursive generator function pow_variants(), awesome actually
Number of divisors (D) of every number (N) can be calculated as:
D = (pow_0 + 1)*(pow_1 + 1)* ... *(pow_n + 1) (1)where
pow_{i}are powers of prime numbers in the source number (N) factorization (N = prime_0**pow_0 + prime_1**pow_1 + ... + prime_m**pow_m). Our task is to minimizeNwith fixedD, so we should check all possible variants of powers which correspond to (eq. 1) and use the minimum possible prime number for maximum power.pow_variants- returns all possible variants of powers which correspond to (eq. 1) sorted in descending order.math.prod(PRIMES[i]**p for i, p in enumerate(powers))- returns minimal possible number for current powers combinationcan you explain please?
yes
This comment is hidden because it contains spoiler information about the solution
learnt this from a previous kata. Glad I remembered it and got to use it.
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For some reason I did not think of this. I feel humbled.
For some reason i assumed the array would be longer than 3 values
FYI: You could replace the implementation of
_to_binwithreturn bin(number)[2:].rjust(INT_LEN, '0').My solution as well...
I think this is clever but it's not something I would leave without comment or in production code