@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 . <[email protected]> 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 <[email protected]>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 <[email protected]> wrote: >> >>> hey i'm also thinking n + logn -2.. but couldnt able to figure out >>> how??? can you please explain the logic >>> >>> -- >>> You received this message because you are subscribed to the Google Groups >>> "Algorithm Geeks" group. >>> To post to this group, send email to [email protected]. >>> To unsubscribe from this group, send email to >>> [email protected]. >>> For more options, visit this group at >>> http://groups.google.com/group/algogeeks?hl=en. >>> >>> >> -- >> You received this message because you are subscribed to the Google Groups >> "Algorithm Geeks" group. >> To post to this group, send email to [email protected]. >> To unsubscribe from this group, send email to >> [email protected]. >> For more options, visit this group at >> http://groups.google.com/group/algogeeks?hl=en. >> > > -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
