11
Last seen 1 year ago
Member for 11 years, 6 months, 28 days
Difficulty Normal
This demonic solution is causing me a frighten!!
Borf is also called on line 7 up to 4999 possible times. Borf is recursive and will generating Fibonacci sequences until it's t / 45 * 3 in size. Every recursive call to borf causes a new list to be created and stays in memory.
Borf has the potent
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That's a pretty neat way to handle that. Didn't think about slicing. I'll have to keep that in mind.
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Nice. wasn't sure if ** would do the trick. Ended up using the math module.
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I know there is a better way to do this. It's quit'n time so this will have to do. :P
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Going out on a limb here... Did you come from a c background?
Nice solution and aptly categorized.
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Wow, that's quite a bit shorter than mine. Nice work. I'll have to look into that frozen set a bit more.
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Nice, pretty much what I did. I ended up using
from datetime import date
and then
abs((date(*date1) - date(*date2)).days)
Just a smidgen shorter but doesn't really seem to make a difference.
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It gets the job done.
Could have moved the cons==3 check under the word.isalpha check that way it would have only run under the condition that it could actually be true. Here it runs even if cons==0. It's really a minor tweak but could help if you're running a really large strings through. Nice wo
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That, is an interesting way to do this. I was trying to figure out a way to pair my solution down a bit. Nice work.
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pop() wasn't really cutting it. It's default behavior is to remove the top item. You can tell it what one you want it to pop by providing it an int but couldn't get it working correctly. Seemed to much fuss to re-porpoise that rather than just use .remove. ;)
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I have been trying to get this paired down a bit but haven't had much luck. Taking a break and will look at it more later.
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Interesting way to do this. Seems like there there is always another way to do stuff in python. :)
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