@dave
akash said "
>Can you please tell me why you chose a bst, simply traversing the array and
incrementing the count would
>have also worked in this case and it would have been O(n).
"
so .. if the range will be high suppose there r few numbers which are very
large so taking an auxillary array and incrementing count will waste a lot
of memory....

On Fri, Jul 2, 2010 at 11:13 PM, Dave <[email protected]> wrote:

> @Jalaj. I don't understand how your comment contributes to the
> discussion. Please explain.
>
> Dave
>
> On Jul 2, 10:56 am, jalaj jaiswal <[email protected]> wrote:
> > for count suppose if there are very big numbers ..then space will nt be
> > efficient
> >
> >
> >
> >
> >
> > On Fri, Jul 2, 2010 at 7:05 PM, saurabh gupta <[email protected]> wrote:
> > > @Dave,
> > > for 2n+1 you can have a configuration where repeated nos may not be
> > > adjacent
> > > you need a block of 3 instead of 2.
> >
> > >  On Fri, Jul 2, 2010 at 6:04 PM, Dave <[email protected]> wrote:
> >
> > >> For problem 1, finding a number that is repeated just once is enough.
> > >> Scan the array to see if there are two adjacent numbers that are
> > >> equal. If so, that is your repeated number. Otherwise, look for the
> > >> repeat in the first 5 numbers. O(n).
> >
> > >> Dave
> >
> > >> On Jul 1, 11:43 am, sharad <[email protected]> wrote:
> > >> > 1.an array of 2n+1 elements is given .....one element is repeated n
> > >> > times
> > >> > and rest all are different.....find the no repeated.
> > >> > 2.same question as above but this time other no's are not
> > >> > different ..i.e
> > >> > they can repeat.
> >
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> > Jalaj Jaiswal
> > +919026283397
> > B.TECH IT
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-- 
With Regards,
Jalaj Jaiswal
+919026283397
B.TECH IT
IIIT ALLAHABAD

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