m getting  result in 95 matches
Ashima
M.Sc.(Tech)Information Systems
4th year
BITS Pilani
Rajasthan




On Mon, Sep 19, 2011 at 7:07 PM, malay chakrabarti <m1234...@gmail.com>wrote:

> 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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