undefined | Better HN

0 pointsabstractbill17y ago0 comments

You just shifted the overhead from removal to insertion. Having different sized notes removes the overhead entirely.

0 comments

5 comments · 3 top-level

sp33217y ago· 2 in thread

I like your idea, but sorting on insertion is O(log(N)), which is faster than searching an unordered list and just as fast as having different-sized bills.

wooster17y ago

With different sized bills:

* Sorting is O(N) [see 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)).)

sp33217y ago

I 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)).

run4yourlives17y ago

How many bills do you carry?

I live in Canada, the land of the debit card. It's rare for me to actually even carry cash anymore.

axod17y ago

Also the fact that the smallest note in the UK (£5) is approx 10 times the value of the smallest note in the US ($1) helps enormously.

Seriously, what's the point of a $1 note? To irritate people everywhere trying to feed them into vending machines?

j / k navigate · click thread line to collapse