@ankit, sorry i was mistaken its O(nlogn) for searching the two elements...
On Mon, Jun 27, 2011 at 8:04 PM, Swathi <[email protected]> wrote: > Dave, > > Can you provide the psuedo code for this.. > > Thanks, > Swathi > > > On Mon, Jun 27, 2011 at 7:30 PM, Dave <[email protected]> wrote: > >> @Sunny. Mea culpa. You are correct. Revised (and correct) algorithm. >> Do two inorder traversals, one in the usual (descend to the left >> before descendung to the right) direction and the other in the >> reversed(descend to the right before descending to the left) >> direction. Let u and r be the current nodee of the two traversals, >> respectively. If u + r < x, then advance the usual traversal and >> repeat the comparison. If u + r > x, advance the reverse traversal and >> repeat the comparison. If u + r = x, and if u != r, then terminate >> with success. If u = r, then terminate with failure. >> >> Dave >> >> On Jun 27, 7:53 am, sunny agrawal <[email protected]> wrote: >> > @Dave >> > i think your solution won't work >> > consider inorder traversal of a BST is 1 6 7 8 15 and x = 14 >> > initially both u,v (1,1) >> > according to u your algorithm will proceed like >> > (1,1) -> (1,6) -> (1,7) -> (1,8) -> (1,15) -> (6,15) ............ -> >> (15,15) >> > >> > and clearly in second step of your solution if (u+v) > x after advancing >> u >> > still u+v will be greater than x >> > so something is wrong >> > I think your solution will work in case we need to find 2 nodes with >> > difference x. >> > >> > correct me if i am wrong.!! >> > >> > >> > >> > >> > >> > On Mon, Jun 27, 2011 at 6:13 PM, Dave <[email protected]> wrote: >> > > @Nishant: No need to store the data in an array. Do two inorder >> > > traversals simultaneously. Let u and v be the current nodes of the two >> > > traversals, respectively. If u + v < x, then advance the "v" >> > > traversal. If u + v > x, advance the "u" traversal. >> > >> > > Dave >> > >> > > On Jun 27, 3:40 am, Nishant Mittal <[email protected]> >> wrote: >> > > > do inorder traversal of tree and store values in an array. >> > > > Now find pairs by applying binary search on array.. >> > >> > > > On 6/27/11, manish kapur <[email protected]> wrote: >> > >> > > > > -- >> > > > > You received this message because you are subscribed to the Google >> > > Groups >> > > > > "Algorithm Geeks" group. >> > > > > To post to this group, send email to [email protected]. >> > > > > To unsubscribe from this group, send email to >> > > > > [email protected]. >> > > > > For more options, visit this group at >> > > > >http://groups.google.com/group/algogeeks?hl=en.-Hide quoted text - >> > >> > > > - Show quoted text - >> > >> > > -- >> > > You received this message because you are subscribed to the Google >> Groups >> > > "Algorithm Geeks" group. >> > > To post to this group, send email to [email protected]. >> > > To unsubscribe from this group, send email to >> > > [email protected]. >> > > For more options, visit this group at >> > >http://groups.google.com/group/algogeeks?hl=en. >> > >> > -- >> > Sunny Aggrawal >> > B-Tech IV year,CSI >> > Indian Institute Of Technology,Roorkee- Hide quoted text - >> > >> > - Show quoted text - >> >> -- >> You received this message because you are subscribed to the Google Groups >> "Algorithm Geeks" group. >> To post to this group, send email to [email protected]. >> To unsubscribe from this group, send email to >> [email protected]. >> 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 [email protected]. > To unsubscribe from this group, send email to > [email protected]. > For more options, visit this group at > http://groups.google.com/group/algogeeks?hl=en. > -- ...Thanks Bharath -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
