oh my god

because 'true' is '1' you can do this. Clever!

That is a very clever solution.

The math you described is qute complex and it is hard to attest it to the problem.
The number of point on each spiral circle is N[k] = N[k-1] * 4 + 4 and that is not mathematical progression.

Total points in whole box is (2*k+1)^2 where k is the spiral sequentel number starting from 0

Your aim is to calculate (2*(k-1)+1)^2 and thereafter calculate how far the movements pass behind this point.

Each side of square we are looking at having size 2*k+1

;)

you literally built a spaceship for solving this kata

Thanks!

Amazing solution, when it's false then result won't count this way ... very nice and clever one.

Haha

bruh!

Леня? о_О

Love it. Just picking up C++ but very satisfying to see (what I feel is) some exemplary code on an 8kyu!!

Согласен)
result.reserve(numbers.size()); после создания добавить

Можно оптимизировать, создав "result" сразу размера "numbers.size()", чтобы не выполнялись лишние медленные перевыделения памяти.
(It can be optimized by creating "result" with exact size "numbers.size()" - this prevents slow memory reallocations.)