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Valentin Bryukhanov
http://bryukh.com/
Last seen 2 years ago
Member for 11 years, 7 months, 16 days
Difficulty Normal
Developer, engineer and taskMaker.
Lovely prints-Alexis_Bourgoin
And you can use triple quotes and make this without \n
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"l" is not good name for variables - because it looks like "1" or "I" for many fonts and hard readable.
if l % 2 != 0 :
can be simpler:
if l % 2:
Your code is good, but PEP8 and good variables name can make nice code from it.
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Nice code. You can make a micro-optimisation with
letter_count = text.count(letter)
With that you don't need count this twice.
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copyOfData = data
For lists it's not a copy, you just created one more link for the same array. If you will change one, then you will change other.
You can count elements with the method "count"
[1, 2, 3, 1, 1].count(1) # return 3
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Very ugly solution , I wish I can do better-ahmedaswai
Whitespaces can be useful. PEP8 can make your code is more readable.
And "Counter" here is overkill, I think. But I sure you can make this pretty and more simpler.
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Sometimes you are using spaces around operators, sometimes - not. Not bad and if you will make it more "general" then it will be nice solution.
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Dijkstra (no fib-heap => O(n^2))-MorrisFeist
Why do you use class here? They look like simple dictionaries here.
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Touch the right wall-maurice.makaay
1
Not universal, but funny :-) And I like "properties". Nice!
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First-undead404
1
Change input data is not a good practice. You can use
work = sorted(data)
instead line 6-7
amount = len(data)
line 9 is overhead
if amount % 2: # the same as amount % 2 != 0
and line 13 is not needed.
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A*-PositronicLlama
Do you think A* is not overkill algorithm for this task? I don't sure about heuristic.
But +2 for implementation and +1 for good comments.
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any([ch.isupper() for ch in data]) and
I prefer list comprehension. For memory problem, i can just to replace list comprehension to generator comprehension. If honestly, i don't like map and filter.
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" == True" is redundant.
"range(0, len(list_of_words))" -- 0 is default, so you can use range(x) == range(0, x)
And you can iterate through list instead indexes:
for w1 in list_of_words:
for w2 in list_of_words:
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