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Awesome Team
Alexander Lyabah
http://www.checkio.org/user/oduvan/
Last seen 13 hours ago
Member for 13 years, 10 months, 19 days
Difficulty Easy
love it!
so many likes. I hope people will also check [our article about f-string](https://py.checkio.org/blog/pep-498-f-strings-python-36/).
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"Harder, Better, Faster, Stronger" - Daft Punk
"Harder, smaller, shorter" - Veky :)
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1-liner: +marriage-przemyslaw.daniel
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Just today I was thinking.. maybe we should make one more funny mission? :)
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If I get this code for a production solution, I would ask to refactor it.
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a video code review from @veky
https://www.youtube.com/watch?v=bAea_X5cCJo
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[Here](https://py.checkio.org/mission/is-even/publications/eugene100372/python-3/first/share/e2ea5c7705f3512261cb241b44526770/#comment-116766) I've explained how `%` works
Operator `not` first converts int to bool...
* 0 = False
* any non-zero value = True
... then inverts
* True = False
* Fals
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A case where solution to the puzzle is harder than the puzzle itself :)
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solution's name looks like a new programming language. You can call it CiO-Spoiler :)
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we should make a video code review around Funny Addition some time :) should be interesting :)
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good solution, and very good direction of thinking.
couple notes here. If you want to check - if list is empty or not, you can simply do:
if not list_to_check:
# list is empty
return False
also this code
if (len(set1) == 1):
return True
else:
return F
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First-viktor.chyrkin
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Imagine how much easier it could be to read.
enemy_mark = 'X' if your_mark == 'O' else 'O'
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Binary count-AleksandrKasianov
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I've just found out that definition of `str.count` is different than `list.count`
>>> help(list.count)
count(...)
L.count(value) -> integer -- return number of occurrences of value
>>> help(str.count)
count(...)
S.count(sub[, start[, end]]) -> int
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