Q9 - > 1 *logn for getting the minimum element ..Normal heap Sort procedure
Q3 - > n+logn-2 comparisions so 51 -2 + log 51 Regards Ankur On Mon, Sep 19, 2011 at 7:59 PM, Ashima . <ashima.b...@gmail.com> wrote: > 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 >>>>>> >>>>>> -- >>>>>> 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. >>>>> >>>> >>>> >>> -- >>> 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. >> > > -- > 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.
