how do you test for time?

great solution bro

i agree. it kind of kills the point of this kata

It has to be one or the other, since all the other values are equal to each other.

I run your in 600+ ms and mine in 260 ms.
I test every value until I find the unique where you sort the whole array.

@kjanus-thecreator Description says. "The tests contain some very huge arrays, so think about performance"
I insist this 6 kyu kata's solution should be implemented in imperative style, not two-lines declarative one. Otherwise it must be 7 kyu.

Isn't this inefficient for large arrays?

This is not an optimal solution, neither space nor time wise. It's not necessary to sort the arrays or even create any sort of counter list / hash. Instead, its possible to do it with four variables, two to store the different numbers and two to store the counts. A more efficient solution only needs a single O(n) traversal of the entire array at worst, as well as O(1) extra space. Dunno why this is so upvoted but to each their own I guess.

Says who? This is a perfectly functional soulution. No need to impose arbitrary limitations on ourselves

All numbers are equal except for one.

this code is not work when you try new double[] {0, 1, 0, 2 ,2}

In this kata, sorting is not appropriate.

This code is just nice to look at.
No sorting required and Low accuracy is problem.
If unique value is most high or low in array, this code maybe good.