@Saurabh. Checking three adjacent numbers won't work, as the example
1,2,3,4,1 shows.

You apparently missed the sentence, "Otherwise, look for the repeat in
the first 5 numbers." If you don't find two equal adjacent numbers,
then there will be a repeat within the first five numbers. How do you
know this? If there are no equal adjacent numbers, then within the
last 2*n-4 numbers, there can be no more than n-2 occurrences of the
repeating number. Thus, the remaining two or three occurrences must be
within the first five numbers.

Dave

On Jul 2, 8:35 am, 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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