i have explained :)

On Sun, Sep 18, 2011 at 11:53 PM, Ashima . <ashima.b...@gmail.com> wrote:

> @malay: how cm n+logn-2?
> cn u explain the logic ?
>
> Ashima
> M.Sc.(Tech)Information Systems
> 4th year
> BITS Pilani
> Rajasthan
>
>
>
>
> On Sun, Sep 18, 2011 at 11:07 AM, Ashima . <ashima.b...@gmail.com> wrote:
>
>> rite! 62.5%
>>
>> Ashima
>> M.Sc.(Tech)Information Systems
>> 4th year
>> BITS Pilani
>> Rajasthan
>>
>>
>>
>>
>> On Sat, Sep 17, 2011 at 9:04 PM, malay chakrabarti <m1234...@gmail.com>wrote:
>>
>>> create a tournament tree.in each round one value is eliminated to obtain
>>> in the process the winner or the highest value in n-1 comparisons. Then
>>> check the queue of the winner which contains log(n) entries of the values
>>> beaten by the winner which implicitly will contain the runners up.Then
>>> log(n)-1 comparisons to find the highest among all the losers whom the
>>> winner had beaten. So all over complexity will be n-1 +log(n) -1 =
>>> n+log(n)-2. Hp that answers ur query. nice question btw :)
>>>
>>>
>>> On Sun, Sep 18, 2011 at 8:02 AM, VIHARRI <viharri....@gmail.com> wrote:
>>>
>>>> hey i'm also thinking n + logn -2.. but couldnt able to figure out
>>>> how??? can you please explain the logic
>>>>
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