This one was tricky for me too. you may be overthinking this. To answer you header question, while loops and recursion are loops that will be faster than a for loop if I am not mistaken. As for the problem, without giving you the answer, (I am bad with hints) try reading the question like this, " how many numbers does "n" have odd and even" (maybe that will help, i know it triggered my brain)? meaning, if we split the number in half we will get 12 even and 1/2 odd (not fully but close) so if we take the number 3 for easy math (haha) 3 = 1 even (2) and 1 odd (1),
do the same thing with 7, 7= 3 even (6,4,2) and 3odd(5,3,1) and 15 = 7even (14,12,10,8,6,4,2) and 7odd (13,11,9,7,5,3,1) but if we take the number "n" and divide it in half we will end up .5 at the end ie (n=15 => 7.5) from there you will want to look at Math Method to achieve the correct answer. I hope I helped and didn't confuse you. your final answer should be one line

## re: What is faster then a FOR..Loop in JS? VIEW POST

FULL DISCUSSIONThis one was tricky for me too. you may be overthinking this. To answer you header question, while loops and recursion are loops that will be faster than a for loop if I am not mistaken. As for the problem, without giving you the answer, (I am bad with hints) try reading the question like this, " how many numbers does "n" have odd and even" (maybe that will help, i know it triggered my brain)? meaning, if we split the number in half we will get 12 even and 1/2 odd (not fully but close) so if we take the number 3 for easy math (haha) 3 = 1 even (2) and 1 odd (1),

do the same thing with 7, 7= 3 even (6,4,2) and 3odd(5,3,1) and 15 = 7even (14,12,10,8,6,4,2) and 7odd (13,11,9,7,5,3,1) but if we take the number "n" and divide it in half we will end up .5 at the end ie (n=15 => 7.5) from there you will want to look at Math Method to achieve the correct answer. I hope I helped and didn't confuse you. your final answer should be one line

developer.mozilla.org/en-US/docs/W...

## Heriberto Garcia Ribas

HGARR-GitHub

OMG of course! I was overthinking it indeed.

There is no need for a loop.

Math.floor(n/2)From now on I will take on the Kata's with a different perspective!

Thanks Jeremy! :)