This is the only way to do it my fellow coder ( ͡° ͜ʖ ͡°)

Google search

can someone explain how this works exactly??

funny how 4 others did it as well

Now it is mentioned.

Thanks!

best solution for sure!

Not only it's slow but this creates a new list instad of removing stuff from the original as stated in description

this invalid code can pass

Are you sure of your solution?
Here three cases where your solution is off by one day compared to http://www.aavso.org/jd-calculator.
(3000, 5, 2) -> 2816909 should equal 2816910 - 2816909 is correct according to the linked calculator
(3001, 1, 7) -> 2817159 should equal 2817160 - 2817159 is correct...
(2100, 6, 4) -> 2488224 should equal 2488225 - 2488224 is correct...
Had to "attempt" several times before getting my solution approved by your random tests.

it's not even clever... ;)

Especially bad since it received best practices instead of clever

I got the impression from the reading I did ( had to do ) that we're being asked for a Julian Day, not a Julian Date. I may be wrong there; if anybody actually knows either way, please chime in.

I didn't know about Julian Days. The description should have contained enough information ( and not necessarily more ) that I would not have had to Google "Julian Date" and read up, esp. the fact that a Julian Day is just a sequential number since a reference date. If you want or need to know more, it's acceptable to have to click through ( a link to Wikipedia would be nice ).

This would be an 8 kyu kata in JavaScript but for the vagaries of Date. I don't know if it is your intention to have this kata being more about pointing out those vagaries than about Julian dates, in JavaScript. It's not necessarily a problem, I don't really know what to do about it either if you think it is, but it's exceedingly simple to convert dates if you know how to encode them in JS.

O(n²), definitely bad.