Nice solution. Reduces number of comparisons to half of Ashish's algo. The complexity remains O[n] though. Can it be made more efficient, like O[log n]
On Thu, Aug 5, 2010 at 10:59 PM, Pramod Negi <[email protected]> wrote: > 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 ... <[email protected]> wrote: > >> In constant space??? How? will you please elaborate? >> >> >> >> On Thu, Aug 5, 2010 at 9:29 PM, dinesh bansal <[email protected]>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 [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]<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.
