Idea is simple either to keep a count and emit elements with odd count (frequency) or XOR element with itself for each time it occurs in input and then emit elements which have non zero XORed result(which basically corresponds to elements with odd frequency).
_dufus On Aug 23, 8:44 am, Nagendra Kumar <[email protected]> wrote: > How are u doing with xor. Can u post ur thought here. > > Thanks > Nagendra > > > > On Sun, Aug 23, 2009 at 2:07 AM, Dufus<[email protected]> wrote: > > > We can count or XOR but I couldnt find any advantage of XORing except > > for preventing overflow. > > > _dufus > > > On Aug 22, 5:03 pm, Nagendra Kumar <[email protected]> wrote: > >> I think you are trying sorting and then counting the frequency of the > >> numbers. > > >> -Thanks > >> Nagendra > > >> On Sat, Aug 22, 2009 at 11:21 AM, Dufus<[email protected]> wrote: > > >> > I can think of a naive algorithm which takes O(n) time and O(n) space. > >> > or O(nlogn) with O(1) space. > > >> > May be someone else might come up with a better algo. > > >> > _dufus > > >> > On Aug 21, 3:01 pm, nagendra kumar <[email protected]> wrote: > >> >> Given an array of integers,Print the integers whose appareance are in > >> >> odd times. > >> >> Need not worry abt order while printing the output. > >> >> Need Algotithm in o(n) time complexity. > >> >> Need efficient space complexity. --~--~---------~--~----~------------~-------~--~----~ 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 -~----------~----~----~----~------~----~------~--~---
