This is good. Another way is to build an interval tree and then query it for every appointment's start time.
On Dec 4, 2:56 pm, Dave <[email protected]> wrote: > If you expect a lot of conflicting appointments: > > Sort the appointments into ascending order by starttime. > Establish an empty min heap of {starttime,endtime} ordered by endtime. > Process the given appointments one by one: > While the heap is not empty and the endtime of the root does not > exceed the the starttime of the appointment > Delete the root of the heap. > If the heap is not empty, then > The given appointment conflicts with every appointment in the > heap. > Insert the given appointment into the heap. > > Dave > > On Dec 3, 10:43 pm, Prims <[email protected]> wrote: > > > > > You are given 'n' appointments. Each appointment contains startime and > > endtime. You have to retun all conflicting appointments efficiently > > > starttime and endtime can range from a few min to few years.- Hide quoted > > text - > > - Show quoted text - -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
