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
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).
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!
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.
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
Bitwise and with 0x1. If result is 0, it’s even. Least significant bit is always 1 for odd numbers.
Lol I think the interview pressure got to you.
Ha ha ha, yup… whoops!
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 }Yep! I’m wrong. Pretty embarrassing!
That’s a nice solution though! Gonna have to try and remember that one!
I’m not following.
n == abs(n)only tells you if it’s positive.yes, I’m being a dumb tonight. I’m completely wrong
Haha, s’all good, yo.