@sagar This is what Dave is suggesting in a more pseudocode way:- 1->Traverse a pointer right down to the leftmost element,i.e.the shortest,say small 2->traverse a pointer left down to the rightmost element i.e.the largest.say large while(small!=large) 3->Compare their sum.If sum>k set large to its successor in reverse inorder.(I am not sure if u meant the same but I am assuming rev inorder to be right->node->left) else set small to its inorder successor. break when u get the desired k. print :) return if u get out of the loop without getting the number then such number does not exist.print :(
Amortized complexity order n. -- Saurabh Singh B.Tech (Computer Science) 5h sem 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.
