This is the famous Inversion Problem.... Hint: you can do it in O(nlogn) using a tweek in merge sort
On Sun, Jun 27, 2010 at 11:26 AM, Anil C R <[email protected]> wrote: > this is just brute force: > > count = 0 > for i in range(0, len(A)): > for j in range(i+1, len(A)): > if A[i] > A[j]: > count += 1 > > Time complexity = O(n**2) > Anil > > > > On Sun, Jun 27, 2010 at 9:59 AM, sharad kumar <[email protected]>wrote: > >> Give an unsorted array find count of pairs of numbers[a,b] where a > b >> and b comes after a in the array. >> >> Eg. {8,3,6,10,5} >> >> the count of such numbers is 5. i.e. (8,3), (8,6), (8,5), (6,5) and (10,5) >> >> >> -- >> 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.
