WTF

nice job using only one counter instead of two.

i sreached it and it very smart way to do it . nice one

My first solution to C++ kata that is the same as any other solution <3

Lol nice, had no idea this existed.

i did something similar but using log10(x)+1 to get the digit amount, now I feel dumb for not thinking of this

std::map is actually O(logn) to find. A hashmap (std::unordered_map) would be what to use for efficiency

very inefficient space-wise.

You are genius!

ascii size is known(128), so u can get the occur-times and put them at their natural-locations in a vector, so that getting the occur-times data for every character only takes constant-level of time.

I'm not an expert but, IIUC, there are more efficient solutions here. This one is O(n log n) due to sorting. But there are some that use either std::map or vectors to keep counts of each letter in each string. Those should be O(n).

excellent

NB!

Many thanks!