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Василий Чибиляев
Last seen 11 months ago
Member for 6 years, 10 months, 20 days
Difficulty Normal
feel free to contact me at chibiliaev23@gmail.com
numpy.rot90 -v1-jianjiaowang
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Great! Though I would consider using list(i) instead of [j for j in i]. And you actually don't need both else statements in the last loop as np.rot90 should be done at each iteration.
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Everything seems to be ok, though i prefer using for loop insted of while
wnenether it is possible
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Easy-solution-nakanohito_piyo
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Very smart. This piece '(i.endswith(j) or j.endswith(i))' does make sense.
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First-Ermakovartem97
Nice and clean. Though I would recommend to replace array[len(array)-1] to array[-1]. Python allows you to use negative indexes to iterate backwards through array, so array[-1] gives you the last element in the array, array[-2] - second last element, ..., array[-len(array)] - the first one. Consider
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Nice, pretty fast and out of the box approach. Well palyed. However I would consider changing this piece of code:
def expand_water(area, mapdata):
for p in area:
l = [q for q in adjacents(p, mapdata)
if is_water(q, mapdata) and q not in area]
area.extend(l)
I
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3-liner: Harder, Better, Faster, Stronger -przemyslaw.daniel
one extra '+' which goes from my heart for the daft punk) oh, the code is awesome!!
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Really great work! You can check out my solution if you want, it is pretty fast but not that complex. Unfortunately I have no opportunity to compare the speed of my code vs. yours, so if you don't mind please do this on your PC and share the results.
[My solution](https://py.checkio.org/mission/mak
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if len(matr) == 2:
return matr[0][0] * matr[1][1] - matr[0][1] * matr[1][0]
these two lines are actually redundant. You can start right from: if len(matr)==1...
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Ok - that was fun. -Olivier_de_La_Fere
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Well, that's pretty creative. You implemented for-else trick and that's great you know such things! Though I don't see big use of it, you get +1 for this. However there are some problems. First of all, your solution is not that flexible. Imagine the input is now 100*100 grid, how many time will it t
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