compare pair wise i.e a[0] and a[1]....a[2] and a[3].. leave out last 4 elements...
repeated element can be found in 3 comparison for these 3 elements... total no of comparison in worst case (n/2+1) Negi On Thu, Aug 5, 2010 at 10:55 PM, Shiv ... <shivsinha2...@gmail.com> wrote: > In constant space??? How? will you please elaborate? > > > > On Thu, Aug 5, 2010 at 9:29 PM, dinesh bansal <bansal...@gmail.com> wrote: > >> If I understand the question correctly... there is an array of size n >> which has n/2 distinct elements and one element is repeated n/2 times. >> >> For e.g.: >> n = 4: 1 2 3 3 >> n = 6 1 2 3 4 4 4 >> n = 8 1 2 3 4 5 5 5 5 >> >> So now this problem can be reduced to finding the first duplicate element >> in the array because remaining other elements will be unique. I think this >> can be done in linear time. >> >> >> -- > 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<algogeeks%2bunsubscr...@googlegroups.com> > . > 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 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.
