3 ms·
Nitpick on Nitpick on Nitpick, it is indeed O(2^m) and not O(e^m). Those are not the same complexity. Logs get to not care, but when we exponentiate it matters
by thethirdone 7y ago
Nitpick on Nitpick on Nitpick, it is indeed O(2^m) and not O(e^m). Those are not the same complexity.
Logs get to not care, but when we exponentiate it matters.