log n is impossible. the other solution i thought was of building a tree where each node contains value and its count. and then building a heap out of these counts, but this will be overkill.
the fact that rest of the n/2 elements are not unique is the killer in the logic otherwise only n/2+1 elements are sufficient Best Regards Ashish Goel "Think positive and find fuel in failure" +919985813081 +919966006652 On Thu, Feb 9, 2012 at 2:53 PM, atul anand <[email protected]> wrote: > i guess can be done using binary indexed tree. > > > On Thu, Feb 9, 2012 at 2:07 PM, Prakhar Jain <[email protected]> wrote: > >> But in this post, we don't have prior information about what can be >> possible majority element. >> >> According to my question, we know that either x is the majority element >> or there is no majority element. >> Can we use that information to reduce complexity to O(log n)..??? >> >> -- >> 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. > -- 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.
