code for my *second *solution http://www.ideone.com/oxDql <http://www.ideone.com/oxDql>Point out any bugs if you find.
On Sat, Jul 9, 2011 at 12:43 PM, saurabh singh <saurab...@gmail.com> wrote: > > Sort and take window of 2 while traversing, > stop when the elements in the window do not match.The first one will be the > non-repeated assuming only one such number exist in the array. > complexity:o(nlogn) > > PS:Another possible solution is => > 1. Form a BST.Add extra variable count to the node. > 2.while inserting check if value already exists,if exist increment count.Do > not insert the duplicate value. > 3 perform any traversal to look for node with count field 1. > > I am not sure of the correctness or complexity of my latter solution,Kindly > ignore if it sounds crap, > Saurabh Singh > B.Tech (Computer Science) > MNNIT ALLAHABAD > > > -- Saurabh Singh B.Tech (Computer Science) MNNIT 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.
