38
Awesome Team
Stefan Pochmann
http://www.stefan-pochmann.info/
Last seen 1 hour ago
Member for 9 years, 1 month, 7 days
Difficulty Normal
Recent solutions I'm happy with (just starting/trying this):
[Words Order](https://py.checkio.org/mission/words-order/publications/StefanPochmann/python-3/short-dict-subsequence/share/5bbb2df54ec5a810d36d7f70ae7e92da/)
Dang it no markdown here?
Another simple oneliner-StefanPochmann
5
Shortening it by removing code duplication (but I'm afraid it makes it a little less clear):
def create_intervals(data):
return list(zip(*(sorted(x for x in data if x+d not in data)
for d in (-1, 1))))
And golfed:
create_intervals=lambda X:list(zip(*(sort
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Shortest? 52 - Kill me now-StefanPochmann
3
Took me forever to find those index expressions. My progression was:
checkio=lambda n:'Fizz Buzz'[(n%3>0)*5:4+5*(n%5<1)]or`n`
checkio=lambda n:'Fizz Buzz'[n%-3&-4:4+5*(n%5<1)]or`n`
checkio=lambda n:'Fizz Buzz'[n%-3&-4:9-n%5-n*4%5]or`n`
checkio=lambda n:'Fizz Buzz'[n%-3&-4:12-(n%-5&8
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Another simple oneliner-StefanPochmann
3
2
Zip the interval starts with the interval ends. Similar to [my previous solution](https://py.checkio.org/mission/create-intervals/publications/StefanPochmann/python-3/simple-oneliner/), so I could've seen this myself, but really just a corrected and improved version of [Lemmi's solution](https://py.
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1-liner: 62 chars? that's it!-przemyslaw.daniel
2
1
Still one character longer than [mine](https://py.checkio.org/mission/house-password/publications/StefanPochmann/python-3/shortest/?ordering=most_voted&filtering=all) :-P
But it's trivial to make yours shorter. Why don't you use a set comprehension?
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5-liner: bit by bit-przemyslaw.daniel
2
Dictionary version :-)
```python
def calkin_wilf(n: int) -> tuple[int, int]:
result = {'1': 0, '0': 1}
for bit in f"{n:b}":
result[bit] = sum(result.values())
return tuple(result.values())
```
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Just shortened the third line a bit (required changing the characters):
checkio=lambda i,r=range:int(bytes(48+min(r(10),
key=lambda n:bin(ord('歫⒚玧碧䧭㣏毎ኧ篯㧮'[n])
^sum(i[o//3][d+o%3]<More
Ah, I should've suspected you've done this already.
No need for `b` if you use `find` instead of `index`.
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simple-StefanPochmann
2
Alternatively, a little more efficient and maybe clearer:
def merge_intervals(intervals):
result = []
for interval in intervals:
if not result or interval[0] > result[-1][1] + 1:
result.append(interval)
else:
result[-1] = r
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I finally [beat you](http://www.checkio.org/mission/fizz-buzz/publications/StefanPochmann/python-27/shortest-52-kill-me-now/)... omg that was hard work. Thanks for the motivation :-)
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Disproof: From goal state, just swap blue with yellow. You can't solve it because you never turn rings 2 and 3.
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For example `correct_sentence('I')` crashes ("I." is a perfectly fine sentence, and since a point of the mission is to correct a missing dot, `'I'` is a perfectly reasonable input).
Btw, how about `tail.strip(".") + "."`?
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A bunch of one-liners-gleb10101010101
1
Uh... That's not a bunch of one-liners, that's one bunch-liner :-)
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Shortest? (54)-StefanPochmann
1
Thanks to [james.verran](http://www.checkio.org/mission/end-of-other/publications/james.verran/python-27/endswithtuple/) and [MadCow234](http://www.checkio.org/mission/end-of-other/publications/MadCow234/python-27/the-end-of-other/) for pointing what `endswith` can do.
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Too complicated and inefficient and unpythonic. This really should not be top-rated.
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2D Recursion (96)-StefanPochmann
1
Readable version:
def checkio(A, i=0):
if not A:
return 1
if i == len(A):
return 0
return checkio(A[:], i+1) + (-1)**i * A.pop(i)[0] * checkio([r[1:] for r in A])
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I'd say removing less than nothing shouldn't remove *everything*, like this does for `remove_min_max({1, 2, 3}, -1)`.
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uno reverse card-viktor.chyrkin
1
1
Inspired by yours:
def remove_min_max(data: set, total: int) -> set:
for extremum in [min, max] * total:
data.discard(extremum(data, default=None))
return data
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