ahh yes, as prakhar says, if the array is bitonic, my approach will work
for O(log n).
On Sun, Jun 24, 2012 at 1:57 AM, Prakhar Jain <jprakha...@gmail.com> wrote:
> I think it can't be done in O(log n) as per given problem constraints.
> It can be done in O(log n) if additional information that "array is
> bitonic" is given.
>
> --
> Prakhar Jain
> IIIT Allahabad
> B.Tech IT 3rd Year
> Mob no: +91 9454992196
> E-mail: rit2009...@iiita.ac.in
> jprakha...@gmail.com
>
>
>
> On Sun, Jun 24, 2012 at 1:45 AM, Sourabh Singh
> <singhsourab...@gmail.com>wrote:
>
>> @adarsh kumar
>>
>> are u sure it's worst case will be O (log n) ??
>> i think iff array is fully sorted O(n) will be required to say "NO
>> such element present"
>>
>> On Sat, Jun 23, 2012 at 1:11 PM, adarsh kumar <algog...@gmail.com> wrote:
>> > This is a variation of binary search, the difference being that we have
>> to
>> > search for an element that is greater than its immediate left one and
>> lesser
>> > than its immediate right one. Just implement binary search with these
>> > additional constraints, thereby giving O(log n).
>> > In case of any difficulty/error, let me know.
>> >
>> > On Sun, Jun 24, 2012 at 1:27 AM, Hassan Monfared <hmonfa...@gmail.com>
>> > wrote:
>> >>
>> >> Given an array of integers find a peak element of array in log(n) time.
>> >> for example if A={3,4,6,5,10} then peak element is 6 ( 6>5 & 6>4 ).
>> >>
>> >> Regards.
>> >>
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