6 ms·
To be precise, we can prove that no sort comparison-based algorithm can do better than O(n log(n)) comparisons. There are some variations though that do better
by davuinci 7y ago
To be precise, we can prove that no sort comparison-based algorithm can do better than O(n log(n)) comparisons. There are some variations though that do better than that, though they are not based on comparisons (i.e counting sort, radix sort).