sorry for the interruption,we can make it work even if the elements are repeated, by removing the duplicacy in linear time(as the array is already sorted) and taking a count of no. of duplicates in the seperate array.
On Wed, Dec 1, 2010 at 9:37 PM, Senthilnathan Maadasamy < senthilnathan.maadas...@gmail.com> wrote: > A small correction to the algorithm above. In Step 3, instead of finding > *any* pair (a,b) such that a+b = x we need to find *all* such pairs. > However the complexity remains the same. > > -- > You received this message because you are subscribed to the Google Groups > "Algorithm Geeks" group. > To post to this group, send email to algoge...@googlegroups.com. > To unsubscribe from this group, send email to > algogeeks+unsubscr...@googlegroups.com<algogeeks%2bunsubscr...@googlegroups.com> > . > For more options, visit this group at > http://groups.google.com/group/algogeeks?hl=en. > -- Anoop Chaurasiya CSE (2008-2012) NIT DGP -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algoge...@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.
