38
Awesome Team
Stefan Pochmann
http://www.stefan-pochmann.info/
Last seen 4 hours ago
Member for 9 years, 1 month, 3 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?
Short, Simple, Efficient, Clear-StefanPochmann
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`a` and `b` keep track of the best sums up until the previous two steps. [One-liner version](http://www.checkio.org/mission/stair-steps/publications/StefanPochmann/python-3/short-simple-efficient/).
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`chi` is nice. But what does the name mean here?
I don't see a reason why this wouldn't work for other sizes (as long as it's small enough to be fast enough).
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Permutations (6 lines)-StefanPochmann
(ok 8 lines but I could easily `__import__`)
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Short, Simple, Speedy-StefanPochmann
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"Speedy" as in I get it after just one guess. I challenge you to find two expressions which are *not* equivalent but which my x,y pair fails to distinguish.
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125-StefanPochmann
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125 is the file size in bytes. It's 127 with double spaces instead of tabs. Here's another 127:
def total_cost(C):
M={}
for c in C:M[c[:10]]=M.get(c[:10],0)-(-int(c[20:])/60)
return sum(m+max(0,m-100)for m in M.values())
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Recursive Cycle, Short-StefanPochmann
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Credit to @jimako1989 for his `all([x+y>z,y+z>x,z+x>y])` which I just shortened to my `a+b>c>b-a>-c`.
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82-StefanPochmann
Alternatively (same length):
checkio=C=lambda c,n,s='',i=0:s*(i==n*2)or C(c,n,s+','*(i>i%min(c,n)<1)+`i%n`,i+1)
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64, using reduce-StefanPochmann
Same as [this](http://www.checkio.org/mission/stair-steps/publications/StefanPochmann/python-3/short-simple-efficient-clear/).
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Can't be bothered...-StefanPochmann
... to write a regex distinguishing the two styles.
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code reuse ftw-StefanPochmann
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Just reusing [my solution to the original mission](https://py.checkio.org/mission/roman-numerals/publications/StefanPochmann/python-3/short-and-clear/).
I could write it as
reverse_roman = list(map(checkio, range(4000))).index
but then I'd have to write it *below* the `checkio` function :-(
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Small-StefanPochmann
Apparently the judge doesn't even care whether I return the intervals as tuples or lists...
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50 (based on przemyslaw.daniel's 62 and artakase's 59)-StefanPochmann
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Apparently the judge doesn't care whether I actually return a `bool`. Here's a 53 that does:
checkio=lambda p:p[9:7+len({ord(x)>>5for x in p})]>''
And here's a 54 based just on przemyslaw.daniel's:
checkio=lambda p:len(p)-7>2>5for x in p})
@przemyslaw.daniel's 62 is [here](h
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Small-StefanPochmann
A variation, one character shorter and perhaps a little nicer/faster:
def merge_intervals(intervals):
r, E = [], float('-inf')
for s, e in intervals:
if s > E + 1:
r += [s, e],
E = r[-1][1] = max(E, e)
return r
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[old] oneliner, should be O(n)-StefanPochmann
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This should be O(n) because the `sorted` shouldn't be necessary. It only is because the judge currently wants the output intervals in a certain order (even though the problem specification doesn't require that). Already reported this judge bug...
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simple oneliner, should be O(n) (see comments)-StefanPochmann
The actual creation of the intervals only takes O(n) time, but the required sorting makes it only O(n log n). Makes me sad :-(
Could use `range(start, start+len(data))` instead of `count(start)` if I wanted to avoid the import, but `count(start)` is much nicer and faster.
A golfed version (would b
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Um... how do you think `start - data[index] == 1` or `start -data[index] > 1` can ever be true?
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What's the point of explicitly creating those two temporary extra lists in the first line? `sorted` is perfectly happy with getting a `set` and it already returns a `list`.
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Reversing `data` seems like an attempt to make it fast, but it's still only O(n^2) because of the repetition of the expensive line 13.
A little experimental confirmation, doubling the input size roughly quadruples the time (I removed your `print` line for this):
```
>>> from timeit import timeit
>>
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