Already posted 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) Thanks Pramod Negi On Sat, Aug 7, 2010 at 7:14 PM, ankur bhardwaj <[email protected]> wrote: > we can do one thing. compare first element with all others. if it matches > with any of them, we have found our number, else leave the first number and > now the required number is available (n/2)+1 times in the rest of the array, > which can be found in O(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]<algogeeks%[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.
