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With different sized bills: * Sorting is O(N) [see spaghetti sort: http://en.wikipedia.org/wiki/Spaghetti_sort http://en.wikipedia.org/wiki/Spaghetti_sort] *
by wooster 17y ago
With different sized bills:
* Sorting is O(N) [see spaghetti sort: http://en.wikipedia.org/wiki/Spaghetti_sort http://en.wikipedia.org/wiki/Spaghetti_sort]
* Retrieval is O(1).
Further, retrieval is O(1) even in unordered sets, so sorting is irrelevant.
(Also, how is sorting on insertion O(log(N))? For each new bill, finding the correct spot is log(N), so you're still looking at O(N log(N)).)
- sp332 17y agoI only meant for inserting one bill, so N is the number of bills you have already. Using a binary search on a sorted list, either to find a specific denomination you want or to find the correct spot for a new bill, is O(log(N)).