39
H0r4c3
● 18 ● 42 ● 52 Leader of the month
9234/ 9695
Horatiu Crista
Last seen 4 hours ago
Member for 2 years, 7 months
Difficulty Normal
Python, Python, Python...

Best reviews / Newest reviews
Numpy-Pavellver 1
Great solution using numpy! I am ashamed of my solution having 38 lines (also, numpy). More
Two versions: Max-heap, Min-heap, 4 lines each-Phil15 1 1
Great solution! Definitely, I will study the heapq module because it's very useful. More
two key sorted-kdim 1
Nice one-liner solution! And the idea of "two key sorted" was excellent! More
5 strings-kdim 1
Clear and very concise! I do not see an approach to make it more concise! More
First-lechrudik 1
OOP for the win! Great solution! More
First-kdim 1
Great solution! For this problem, I didn't think to use isupper, islower, isdigit. I used regex. More
one long string =)-kdim 1
Sometimes, longer means better! :) More
not password.isdigit()-Phil15 1
Great solution! Because I am a fan of regex, I used re.findall('[\D]+[\d]+', password) for the second part. More
match case-kdim 1
Match-Case statement! Python 3.10. Great! More
First-kdim 1
I used this problem to get to the next stage a few times. More
First-Pavellver 1
Great solution! And very short! More
First-Pavellver 1
Great solution! List comprehension for the win! More
re.sub + while-kdim 1
I like regex a lot. For your solution, I searched the explanation here (https://regex101.com/): r'^#+|[^#]#' 1st Alternative r'^#+ r means raw string ^ asserts position at start of a line '#' matches the character '#' + matches the previous token between one and unlimited times, as many times as po More
zip + count-kdim 1
Great approach with '+-'! And only 2 lines. My solution has 30 lines. More
one string-kdim 1 1
The second part could be replaced by: re.findall('[\D]+[\d]+', password) I am a fan of regex. :) More
5 lines - sub + split + eval-kdim 1
A good lesson in regex! Bookmarked! More
classic-kdim 1
Nice use of slicing in list comprehension! More
First-Red_Ale 1 1
Simple and clear! I used re.findall('[\D]+[\d]+', password) for the second part, because I am a fan of regex. More
First-Pavellver 1
Good idea of using enumerate! I converted the zip in set and compared the lengths. More
First-Red_Ale 1
Nice solution! For the second condition, I used regex: re.findall('\d', password) More
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