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 > > -- > You received this message because you are subscribed to the Google Groups > "Algorithm Geeks" group. > To post to this group, send email to algogeeks@googlegroups.com. > To unsubscribe from this group, send email to > algogeeks+unsubscr...@googlegroups.com. > 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 algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
