27
Федор Кудинов
Last seen 10 hours ago
Member for 4 years, 10 months, 5 days
Difficulty Normal
return ing list in line 5 makes no sense since return of generator is not used, you can replace line 5 by:
return
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I believe idea of a task is to take items from iterator 1 by 1 and yield them as you go.
but in line 2 you process all items at once.
Imagine if you receive as input never ending iterable object.
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13 - 16, 24 -27,
hardcode is usually not good idea, I prefer small loops for repeating parts of code
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intervals.pop(0) makes you stick to specific data type i.e. list
better to take iterator from intervals and use next to get 1 item
iterator = iter(intervals)
a0, b0 = next(iterator)
of course StopIteration error can happen if iterator is empty
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to be speedy, you need at least use generator expression
your solution is not memory effective
def sum_upto_n(N: int) -> int:
return sum(i for i in range(1, N+1))
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also if you add 0 stone to stones, you can remove if else block, and leave only 1 return
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better to avoid sorting on every loop,
sort once, and insert into expected position to leave stones sorted
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I would argue that such way will also do the job
((x + 1, y), (x - 1, y), (x, y + 1), (x, y - 1))
but still I saw some practical example of ~ usage, rather tricky one like usage of bitwise shift to multiply 2, but still
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ideally you should not use magic methods
you can take same functions from 'operators' module
from operator import add, sub, mul, truediv
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If you need to work with dates, better to spend time learning how to use python date, datetime
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oh wow so over enjeneered
while loop could easily handle this )
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if to move
len(items). to sepparate variable it will simplify code and suggest ways for further simplification
i think 2nd list could be handled by negative indexing,
smth like:
items[-len(items)//2:]
but need to check of course
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