this post was submitted on 21 Dec 2024
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Fun Loops ▶️

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[–] [email protected] -3 points 2 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!

[–] [email protected] 8 points 2 months ago (2 children)

That would be isPositive.

Without using the modulo operator you'd essentially have to reimplement it. Divide the number by 2 and round down. Multiply that by 2 and then subtract it from the original number.

isEven(10) results in 10-10==0 (true) whereas isEven(13) results in 13-12==0 (false).

function isEven(n){
  n = Math.abs(n)
  return (n - (Math.floor(n/2) * 2)) == 0
}
[–] dream_weasel 4 points 2 months ago* (last edited 2 months ago)

Minor simplification: this works even without taking absolute value first of you use fix instead of floor.

Edit: I don't know if fix is in the stock math library on second thought...

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