L7, from the first array make a hash table H-> O(n) go through the second array and test if the element is present in H1 ->O(n)
so the solution is 2*O(n)=O(n) Your solution: sort the first array -> O(n log n) sort the second array -> O( n log n) compare the two arrays -> O(n) the solution : 2*O(n log n) + O(n) = O( n log n) Siva how will you use the B-tree, so you can make one node to correspond to one block in the disk but how will you contro the writing the each node to the disk. This is a issue of low level programing for me. Siva please explain? --~--~---------~--~----~------------~-------~--~----~ 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 -~----------~----~----~----~------~----~------~--~---
