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Pick one :/
reviewd my code and tested it on all combination - it works 100%. I have an internal error
and it makes sense -> my code fails always on the 15 test - the test are random and I found it fails
on tests which I allready succeeded in previuos attempts. ):
I just solved it myself and it's working fine, except that assertion messages are horrible. Not working code != kata issue.
Please refer to https://github.com/codewars/codewars.com/wiki/Troubleshooting-your-solution for some tips.
The error is coming from your code, not the tests... And if not, you'll have to prove it.
The error is:
TypeError: Cannot read property '0' of undefined
at addPositions
at turg
at /home/codewarrior/index.js:315:29
at /home/codewarrior/index.js:513:5
at Object.handleError
thrown after solving all the tests - the reason I'm not specific is that if I include my code than the message is hidden and I can't see it since it does not let me solve it.
What's the error??? And why is that a kata issue? (I haven't solved it, so I can't see your fiddle)
If you're gonna raise an issue, please BE SPECIFIC.
My solution works for any givven position of the K tested on the site and in jsffidle.
When I submit it throws a js error after passing all the tests -> check the hidden comment with the link to the fiddle
test that iterate all possible cases...
This comment is hidden because it contains spoiler information about the solution
No, your solution isn't "much more efficient". Only one iteration is needed. You still iterate two or three times.
You don't even need to filter the whole array, just the first 3 numbers.
your solution must be top one
It actually takes 3N iterations. The result can be found by the third number, so no real need to loop further.
Sometimes I feel we FE devs get so caught up in functional programming we forget about efficiency of implementation.
Thank you.
The second condition covers n = 0 as 0 % 2 === 0 resolves to true, so n < 0 is OK. It could be n < 1 if you want to make your code a bit "faster".
Should be
n < 1:)Zero is not a diamont 1 is the smallest.
Great Solution!!!
You don't need to filter both every time - Just filter evens and if evens length is greater than 1
then you can filter the odds inside the terinary condition.
Also no need to parse int JS will assume its numbers by it self.... :)
Take a look at my answer its much more efficient