27
Last seen 8 hours ago
Member for 1 year, 6 months, 15 days
Difficulty Normal
Thanks for that solution. I didn't know you could flatten lists/sets like that.
sum(matrix, tuple())
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Yeah, thats the good stuff! I would give you points for it, but sadly that doesn't work in 'random soulution'.. ;)
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Oh wow, now THIS one is quick :D
Takes 0.00 seconds for 1000 runs of is_perfect(100000000). Mine takes 0.40 seconds.. Nice one!
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It's very slow though.. Takes over 4 seconds for n == 100'000'000 on my computer. You could at least go for
divisors = filter(lambda x: n % x == 0, range(1, (n + 2) // 2))
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it's pretty slow though.. Takes 2.5 seconds on my computer for is_perfect(100000000). You could at least go for
def is_perfect(n: int) -> bool:
def find_divisors(m):
for i in range(1, (m + 2) // 2):
if m % i == 0:
yield i
return sum(find_divisors(n))
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Loved your "Clear" solution, but now this is just a nightmare! ;)
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Loved this one! Lots of things to learn here.
code[i] or '0'
Nice!
'01'[i in rs]
Very nice! :)
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No need for those len()'s :D
import re
def checkio(expression):
s = re.sub(r'[^\][(){}]', '', expression)
n = 1
while s and n:
s, n = re.subn(r'\[]|\(\)|{}', '', s)
return s == ''
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I like the logic, but I think you could make it more compact
def checkio(exp):
# From the input exp, create a new string with only the brackets allowed in it.
b_in_exp = ''.join(c for c in exp if c in '()[]{}')
# Replace all brackets in the new string with empty strings. Retur
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Fixed?-StefanPochmann
1
1
Very nice! Thanks for reminding me that map can take more than one iterable.
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No need for the list comprehensions and the len :)
return max(args) - min(args) if args else 0
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if s2[0] >s1[0]+s1[2] or s2[0] < s1[0]:
if s2[1] > s1[1]+s1[2] or s2[1]s1[0]+s1[2] or s2[0] < s1[0]) and (s2[1] > s1[1]+s1[2] or s2[1]More
if s2[0]<=s1[0]+s1[2] and s2[0]>=s1[0]-s2[2]:
return True
else:
return False
could be
return s2[0]<=s1[0]+s1[2] and s2[0]>=s1[0]-s2[2]
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This code could need some cleanups..
For example, all the print statements are redundant. You could replace the 'items[len(items)-1]'-like statements with just 'items[-1].
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I have some issues with this solution..
In combinations_with_replacement you take range(1, 18) -> 1 to 17, but max(die) is <= 18.
Also, you could optimize that to range(1, min(19, sum(enemy_die) - len(enemy_die) + 2)
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No need for an 'else' clause:
from collections.abc import Iterable
def duplicate_zeros(donuts: list[int]) -> Iterable[int]:
res = []
for x in donuts:
res.append(x)
if x == 0:
res.append(x)
return res
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