Right. We did the same problem for n^2 instead of n^3. There is code at
http://groups.google.com/group/algogeeks/browse_thread/thread/80e96343701c0fd9/0509d9f83b0d6ae8 On Oct 3, 9:07 am, vaibhav shukla <vaibhav200...@gmail.com> wrote: > * > > lets have another approach,tell me if it works > > Notice that the number of digits used to represent an n^3 different numbers > in a k-ary number system is d= log(n^3) base k. > > Thus considering then 3 numbers as radix n numbers gives us that: > > d=log (n^3)base n= 3logn(n)= 3 > > Radix sort will then have a running time ofΘ(d(n+ k)= Θ(3(n+ n)) = Θ(n) > > * -- 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.
