Hi Rahul,
In the below url,They have mentioned the parallel searching. it means
divide array than search element from two point.
i.e number of element is {48,23,10,32,5} search 32.
divide array [0-2] and [3-4] range... traverse the array from p[0] and p
[3]... till half of the loop.
I hope we can search element into log n ( need to look more just giving .2
cents)
http://www.medwelljournals.com/fulltext/?doi=rjasci.2011.70.75
On Sun, Jun 3, 2012 at 9:25 PM, Rahul Kumar Patle <patlerahulku...@gmail.com
> wrote:
> @abhinav: if you want to search just 15 in log(n) time then you can use
> the concept of heap tree.. apply one round of heapification (not for all
> elements but just one time it will be complete in log(n) times), and you
> will need to swap elements but when you got element 15 you can stop..
> although space complexity has increased... you will need one redundant
> array to use heap operation so that finally you will have original array as
> it is...
>
> Thanks and Regards:
> Rahul Kumar Patle
>
>
> On Sun, Jun 3, 2012 at 8:31 PM, abhinav gupta <abhinav....@gmail.com>wrote:
>
>>
>> We have given a list 14 6 7 15 8 9 ............we have to find 15 in
>> (log n ) times.
>> --
>>
>> *Thanks and Regards,*
>>
>> Abhinav Kumar Gupta
>> **abhinav....@gmail.com
>>
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