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[R] Stupidest ways people have solved coding interviews (loops.video)

submitted 5 months ago by m_f@midwest.social to c/loops@midwest.social

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[–] Kraiden@kbin.earth -2 points 5 months ago (11 children)

I'm sorry, but even without knowing about the mod operator, this is inefficient and over engineered. Why have a loop at all?

fun isEven(n: Int){
    return n == abs(n)
}

no loop required...

having said that, I can totally see how that was missed in a high pressure interview. I hate interviews like that!

edit: Ha ha... isEven...not isPositive... I'm tired. ignore me!

[–] lobut@lemmy.ca 11 points 5 months ago* (last edited 5 months ago) (2 children)

Because the abs(3) == 3 is true and that isn't even.

An even number of flips would be true and an odd number of flips would be false which works out.

I was thinking a bitwise & or converting it to a string and testing if the right most character is 0, 2, 4, 6, 8 would be panic mode solutions too.

[–] Kraiden@kbin.earth 3 points 5 months ago (1 children)

you might be able to do it with a bitwise op? My track record tonight is not great so I'm not going to comment. Have a look at @ImplyingImplactions comment for a loopless solution

[–] UID_Zero 3 points 5 months ago

Bitwise and with 0x1. If result is 0, it's even. Least significant bit is always 1 for odd numbers.

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