zingbretsen @zingbretsen@social.cleverlibre.org
Data scientist, nerd, Vim & Emacs evangelist.
Joined Sep 2019
@Absinthe how do you feel about importing modules for these challenges? One standard lib module would make the no-division part really elegant.
@masterofthetiger@theres.life @Absinthe
One last style thing is that python has some syntactic sugar regarding mathematical operations so you could do:
output[x] *= n
Which multiplies output[x] by n and stores the value back in outout[x].
A minor thing, but is generally the more pythonic way of doing it
Finished code (Python)
@masterofthetiger@theres.life @Absinthe and style-wise, i is usually used as the index when looping over numbers, so it's confusing that it's storing your values here. Also, since this is something that's so common to do, python provides a way of looping over both the index and the value at the same time: enumerate
for i, elem in enumerate(input):
#do stuff
Finished code (Python)
@masterofthetiger@theres.life @Absinthe
I haven't tried running it yet, but I'm pretty sure you would get the wrong answer if the input list had repeated elements, e.g., [5,1,2,3,4,5]. Because you are checking that n != i, where i is the current element in the input list, if that element occurs more than once, you will not include that in the product. I believe the question wants you to exclude only the element in that given position on the list, not all elements of that value.
Finished code (Python)
@masterofthetiger@theres.life @Absinthe would you mind if I gave you some feedback on your code?
zingbretsen
boosted
Okay, here is another freebie :) I will put in a real one before the end of the weekend.
Basically, you are given a list from which you need to create a new list where each element is the product of all the "OTHER" elements.
I found a few interesting corner cases.
I challenge you to give it a try!
Read the challenge here:
https://git.qoto.org/Absinthe/productnot/blob/master/README.md
My attempt is checked into that repo as well.
@Absinthe @freemo I forked the repo and added my solution here: https://git.qoto.org/zingbretsen/coconuts/blob/master/coco.py
It should solve the 5 person problem when run as-is, but it could be run for any number of people. I should probably have added that as a command line argument.
It works backwards from possible ending amounts, but is somewhat selective in the numbers it tries. e.g., for the 5 person case, we know that the ending pile must be divisible by both 4 and 5, so the final pile will be a multiple of 20. In addition to that, we know that 5/4 + 1 of the ending pile must be divisible by 5, so 60 is the lowest number that would work for the first round of stepping backwards.
There is a regular pattern that seems to work for all of the numbers of people I tested, not just 5, so we can reduce the search space pretty significantly. For the 5 person problem, I only have to test 13 numbers before finding the answer.
Data scientist, nerd, Vim & Emacs evangelist.
Joined Sep 2019
