Hello everyone, I hope you are doing good!
I am not sure if this is the right way to ask a question in here? Forgive me in that case. I just joined the community. But I hope somebody can help me by pointing me in the right direction?
I am sitting at the moment in a coffee bar in The Hague (The Netherlands), not be confused with a coffee shop! ;) I am getting frustrated about a seemingly easy Javascript Kata...grrrr! So maybe a coffee shop would be a good idea at this point :)
The Kata instruction:
Given a number n, return the number of positive odd numbers below n.
Expect Large Inputs!
My first solution was to loop using a simple For..Loop from 0 to n-1:
for (let i=0;i<n;i++){if (i % 2 != 0) count++}
Codewars returned the error
Execution Timed Out (12000 ms)
So I thought to make the loop a bit smarter and not loop through all the numbers but only looping through the odd numbers like this:
for (let i=1;i<n;i+=2){if (i % 2 != 0) count++}
But this is still not fast enough for Codewars.
Googling didn't help me find an answer. So I am hoping somebody here knows what I am overlooking? Or is this a Codewars glitch? I tried several times on different days.
Thanks in advance!
Happy coding!
Heriberto
Top comments (2)
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
developer.mozilla.org/en-US/docs/W...
Heriberto Garcia Ribas
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! :)