No, the k solution computes the unique elements and group independently. #:'=v[;1] uses group for the first column, and ?v[;1] uses unique for the second. (;) i

No, the k solution computes the unique elements and group independently. #:'=v[;1] uses group for the first column, and ?v[;1] uses unique for the second. (;) is list notation. Your analysis of the algorithm you've described is also wrong: it would be O(n*log(n) + n^2), or O(n^2), because the second part loops over unique elements and not operations performed by unique. In the random-access model, hash table construction/lookup is O(n) average case for n elements, if the elements have bounded size. More generally, it's linear in the total size of the input. If the strings are very long it can take a while to hash and compare them, but there's never a log(n) factor. Maybe you're confusing this with results from comparison sorting? Both solutions are O(n) with the assumptions of random-access memory and bounded string size (or hash/comparison time), which I consider to be pretty standard.