@Sagar: Don is correct. n/2+n/4+n/8+... ~= n. But even the first round, involving n/2 comparisons, is O(n).
Dave On Dec 12, 11:23 am, sagar pareek <sagarpar...@gmail.com> wrote: > Yes Mr. DoN > > First round of Tournament sort results in log(n) time for finding largest > no. > > n/2+n/4+n/8........ results n/(2^i) where ^ = power > > > > > > On Mon, Dec 12, 2011 at 8:16 AM, Don <dondod...@gmail.com> wrote: > > No. To find the largest number in an unsorted array requires looking > > at each number, which is order n by definition. > > Don > > > On Dec 12, 10:02 am, sagar pareek <sagarpar...@gmail.com> wrote: > > > Hi every one. > > > > Can we find largest number from an unsorted array in log(n) ? > > > > -- > > > **Regards > > > SAGAR PAREEK > > > COMPUTER SCIENCE AND ENGINEERING > > > NIT ALLAHABAD > > > -- > > 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. > > -- > **Regards > SAGAR PAREEK > COMPUTER SCIENCE AND ENGINEERING > NIT ALLAHABAD -- 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.
