37
Stensen
19 36 53 Leader of the month
8289/ 8695
Last seen 4 days ago
Member for 3 years, 8 months, 16 days
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The Fox!

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Similar Triangles-Elena_Korljukova 2
I liked the idea except the documentation was in Russian so I didn't understand what you wanted to say. More
native_days_between-Jon_Red 1 1
Weel, I won't say I enjoyed reading it but Nice solution so far :D More
Isdigit 2 ways-Striga 1
I would suggest else if you want: **filter** can do the job + the dunder **\_\_len__** list(filter(lambda i: i.isdigit(), list(text))).__len__() Good Luck, See you soon! More
Sum & multiply 2 ways-Striga 1
Oohh That's interesting carry on the hard work! More
Random white spaces-Striga 1
Cool! Carry On the nice work! ;) More
2 ways: count/remove & count check-Striga 1 1
It could be made as one-liner solution: checkio = lambda data: [i for i in data if data.count(i) != 1] The only thing I'd like to say is that (data.count(i) != 1, data.count(i) > 1) operators both do the same condition! I mean: checkio = lambda data: [i for i in data if data.count(i) > 1 More
Stable Sort With Frequency -kunkka71 1
Excellent! use of len & index. More
Next Birthday-JimmyCarlos 1
Really Nice Solution! I liked it Man! More
Zigzag Array-vvm70 1
Best one-liner I've seen for this mission. More
First-vvm70 1
BTW You cam compress Both functions into one and OFC It's gonna be Function recursion But mainly it would like you're flattening the tree object of objects! But I liked the idea of splitting the function so It would be more comprehensive and clear! More
ezy-danersow 1 1
Since you already have the answer you could make parameter's default value: def friday(day, fridays=[4,3,2,1,0,6,5]): return fridays[datetime.datetime.strptime(day, '%d.%m.%Y').weekday()] More
First-vvm70 1 1
I am shocked! I would vote +200 if it was possible! This is best the solution That made my day today HHHH.. More
First-vvm70 1 1
Mine is better than yours! hehe More
First-vvm70 1
Deserves more than +5! You are the best More
First-vvm70 1 1
Why is it on uncategorized I think it is very clear! More
First-vvm70 1
You've done it linearly, I really respect that amazing idea. More
Some shit-danersow 1
It does take a little of time to find all possible paths which consumes more memory, but if you tried to encounter the second element while you pathfind it would be much more effective and less memory-consuming. But I liked your approach since you try to encounter every single match which can be ver More
First-cnkisk 1
I like the way everything is arranged in your code More
First (='ω'=)-Magu 1
Well Done! Nice use of itertools.count although using infinite loops in your code is a bit risky, however, you did perfectly. More
First-GreenhornPro 1
Your 4th line is a little bit stressed, I would recommend the following, Instead of: arry = list(map(lambda i:i.replace(' ', '').lower(), text.split('\n'))) For Example I'd omit all of the map, list and lambda to minimize time complexity to the following: arry = text.lower().replace(' ', ' More
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