On Monday, October 22, 2012, Dipit Grover wrote: > Since the number of inversions are of order n^2 in the worst case, so > should be the complexity of printing them apparently.
It makes sense to some extent but this is no proof. There has to be a better proof for lower bound of complexity for this algorithm. Because I can state similar statement, "Since the number of inversions are of order N^2 in the worst case, so should be the complexity of counting them apparently". But we all know this is not true as we already know O(nlogn) solution. > -- > 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]<javascript:_e({}, 'cvml', > '[email protected]');> > . > To unsubscribe from this group, send email to > [email protected] <javascript:_e({}, 'cvml', > 'algogeeks%[email protected]');>. > For more options, visit this group at > http://groups.google.com/group/algogeeks?hl=en. > -- Aamir Khan | 4th Year | Computer Science & Engineering | IIT Roorkee -- 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.
