@Don nice solution using max heap.it is very simpler to understand On Tue, Sep 4, 2012 at 8:54 PM, Don <dondod...@gmail.com> wrote:
> Constructing a max-heap is O(n*log n) > > However, the problem asked for a solution using tournament method. > If you ignore that requirement, an O(n) solution is trivial, and > doesn't require the additional storage of a heap: > > int first = max(A[0], A[1]); > int second = min(A[0], A[1]); > for(i = 2; i < N; ++i) > { > if (A[i] >= first) > { > second = first; > first = A[i]; > } > else if (A[i] > second) > second = A[i]; > } > > // First and second now contain 1st and 2nd largest values in A > > Don > > On Sep 3, 5:04 am, bharat b <bagana.bharatku...@gmail.com> wrote: > > Construct a max-heap --> O(n).. > > call delete() 2 times .. --> O(logn).. > > ===> O(n) time.. > > > > > > -- > 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.