Im at 129... did you get it tho?

130!!!

132!

135!

I have joined you at 137!

I was stuck here(138) too, How to reduce it into 128.

I was stuck at 138 and couldn't get it down lower. My maths was as concise as it needed to be. I came back after a while and quickly got it down to 128. 138 -> 128, so you might say the "3" became "2".

Same here. I just get down to exactly 138. Have you got progress in the past few months?

Use sys.setrecursionlimit, or rewrite your algorithm to not use recursion, or rethink the appropriate algorithm for this problem ;-)

This Kata is too hard for you, maybe you'll try this one first.

To test your code, call your height function with some large input that satisfy

sometimes
  n <= 80000
  m <= 100000
while sometimes
  n <= 3000
  m <= 2^200

(i.e. you call see how long does height(80000, 100000) take)

There're about 300+ tests in total, you've only passed 8.

Please improve your solution.

"less noise around"? x)

btw:

• wrapping a Counter with a dict is pointless (if not non sense! x) Well in some cases you need it, but clearly not here)
• you should find something even more obfuscating to check the parity. hex value stays waaay too easy to read... ;) (... & 1 does the same, by the way)
• print??

and what's the error message, at the bottom?

Thanks!

Oh, sorry, it works only when seq contains number with odd occurrences.

This is the task though - return the only integer which appears in the array odd amount of times.

Aaaand, I understood nothing from these "benchmarks" (what were you even testing?). I compared them myself before publishing the kumite to see if there was any difference in speed (same 10000 elements long array tested 2000 times), and both solutions completed the task in 1800-2200 ms (although a couple times my code passed the tests in 1400 or 1600 ms, your execution time never dropped below 1800 ms).