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.
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