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 -- 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.
