xor solution is n. Most efficient

Good one! You also are good writing without "!=0" because boolean for "0" would be "False" anyway, and all others would be "True!
I used set, but actually this one is shorter, maybe it's more expensive if talking about productivity...
Thank you!

There will always be only one integer that appears an odd number of times.

Please read the description. That input isn't valid for this kata.

I think the same than you, maybe its correct because they dont put a test in order like you write but its a posibility so for my is not the correct answer

oh that's smart, I didn't think of the set, but I think I did one step better by just making a range of min, max of the seq

This is an elegant and the most pythonic way of solving the problem however it results in a time complexity of O(n^2)/ space complexity of O(1) whereas if someone did a dictionary its only O(n)/space complexity of O(k). It may be great in terms of space however in thousands to millions of test cases in real life scenarios, it would take time but still a good
code though.

boo golfing, make your code easier to read

Ohhhh, my bad, english isn't my native language... Don't understand word odd) Thank you much)

damn i have never seen anyone write this like that lol. why dont you just use O(n^2)?
And wouldnt you rather then say that your dictionary one would be O(n+m) with m <= n?

in most cases it is one line statment , != 0 is redundant

and I thought my code would is "hella slow"

This is a great demonstration of how terrible is codewars ranking system.

At least add buttons "Bad Practices" and "Bad Complexity".

What is currently here teaches people bad things and reinforces wrong opinions about code.