fixed

Yup, that is the correct observation, thank you very much :)

Oh interesting! I like your solution.

some of us havent got to space/time complexity lol

In theory, yes. Bust in many cases the time saved by using a wildcard is more important. Plus, you can limit your sample to, let's say, 100 rows and not worry about how many columns you have there anymore.

See https://en.wikipedia.org/wiki/Fibonacci_number#Closed-form_expression

phi=(sqrt(5)+1)/2 = 2/(sqrt(5)-1) ~= F(n+1)/F(n) for large n. So F(n) ~= F(n+1)/phi, where by ~= I mean "approximately equals". It turns out the round function can be used since F(n+1)/phi will be within 0.5 of F(n).

I think so because the query will be ran for every row, rather than once and then groupped

Same Big-O complexity ( either O(n log w), O(n log n) or O(n^2) ) depending on the complexity of counting / looking up items.

With small arr, there can be no re-allocation.

We can improve that with a pre-allocated array: visited := make([int]*int, len(arr) - 1).

Nothing says the inmutability of age needs to be preserved. Actually it doesn't.

@mreza-dehghani: you can use a slice to preserve the original array: _data = ages.slice().sort(...

so what is your solution??

thanks for your feedback

There's no need for spam, even if your messages are meant to be helpful.

Thanks, I"m pretty new in golang world :) Love your solution!

Yes, I would only use make([]int, 0, n) when the n is known beforehand, otherwise let the append manage allocation. And in the Go production code you may use profiling: https://golang.org/doc/diagnostics#profiling