@Dave Yours only works for a certain subset of all possible powers or 3 doesn't it? So WgpShashank's would be more general?
On 5 December 2011 14:30, Dave <[email protected]> wrote: > @WgpShashank: Yours is an O(log n) solution. Mine is O(1). > > Dave > > On Dec 5, 6:21 am, WgpShashank <[email protected]> wrote: > > @SAMMM have a look > > > > * *solution is to keep dividing the number by 3, i.e, do n = n/3 > > iteratively. In any iteration, if n%3 becomes non-zero and n is not 1 > then > > n is not a power of 3, otherwise n is a power of 3 > > > > check it out ?http://codepad.org/863ptoBE > > > > Thanks > > Shashank > > Computer Science > > BIT Mesrahttp:// > www.facebook.com/wgpshashankhttp://shashank7s.blogspot.com/ > > -- > 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.
