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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>
>
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