38
Awesome Team
Jürgen Strobel
Last seen 22 hours ago
Member for 5 years, 1 month, 11 days
Difficulty Normal
That's a nice showcase of itertools (although I prefer importing the functions directly). I especially like how `avoidable` works.
It's not very efficient though, `filter(satisfied, ...)` checks whole boards created by `product(*places)`. A lot of the search space could be pruned if placing queens
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Yeah that's the right way to do it. I realized too late that *everything* is to be lowercased except the escapes, and that there is no reason to parse and handle the different parts of the URL separately.
Of course, if the path would countain `:80` followed by a non word character this would break,
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`decimal` can be used to do quantization, rounding and formatting of the result too.
Funnily I didn't use it to find the right power as it can be done with integer math easily.
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The search is much better structured now, but I'd still prefer doing it in integer arithmetic.
if number // base ** i ...
or even
if number >= base ** i or not i:
Yes I may need to repeat the last calculation in Decimal and can't show off a walrus, but given int operations are **much**
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In retrospect, the pattern could be simplified to `rf'({_l}{_u})|({_u}{_u}{_l})'`.
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I am wondering if binary search is strictly optimal, or an approximate solution (polynomial) could be used to guide the search ;-)
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This theoretically fast algorithm avoids a lot of repeated additions and multiplications by intertwining creating permutations and their inner products with `coeffs`.
But in practice eschewing itertools.permutations or numpy does not pay off well, compare to [this](https://py.checkio.org/mission/cr
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First-jaunkunal029
Nice use of setdefault(), but there's a collections.Counter class would make this even more elegant.
Also there's a way to avoid the sorting step, max (and min) can take a key= parameter and compare by 2 properties in a tuple at once, one being the count, the other the character itself, and one mul
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O(N)-PositronicLlama
It's also O(n) in space though, and you'd just need to keep the last 2 elements of `values` while iterating, making it O(1) in space.
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OK let me be blunt: The len()==1 case shouldn't be necessary even for your code, as the next case would handle it just fine. Also you could just pick the middle 1 or 2 list elements by indexing [len(data)//2)] instead of removing everything else. I don't see how i is used, it's completely superfluou
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In Battle if an army is empty this will end with an UnboundLocalError since soldier_1_alive is never assigned to. No precondition to assure armies have units.
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Index-based iteration in Battle.fight is non-pythonic (some might even say clumsy). Better options are explicit iterators + next() or popping the dead units off the armies.
Btw catching the IndexError doesn't save you if army u1 is empty, since u1.members[-1] will throw another one.
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I don't see how the key lambda can work? In the first example both 6 and 2 get the same frequency hence the output should be in order of input: [4, 4, 4, 4, 6, 2, 2, 6]. That was my first attempt too but it got rejected.
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First-vitalik_force
You could define j before inline, and use j instead of jj in inline.
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First-igor.v.dudenko
Works, but rather brute force. Can be done in a single loop by adjusting a sliding window's sum up and down.
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Works, but rather brute force. Can be done in a single loop by adjusting a sliding window’s sum up and down.
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