If we used a sort of O(n), then did our checks, which at most is O(n), then I’d say IsIsogram() could be of O(n).
However, my vote is O(n log n), as I’m unsure of how efficient sorts are in C++ yet.
If we used a sort of O(n), then did our checks, which at most is O(n), then I’d say IsIsogram() could be of O(n).
However, my vote is O(n log n), as I’m unsure of how efficient sorts are in C++ yet.