actually the problem is to make BST construction in O(nlogn)... we need rotations which changes the structure so you wont be able to distinguish between elements right to a particular elements, elements left to it . ( in original array )..
On Tue, Jul 13, 2010 at 12:52 AM, Tech Id <[email protected]> wrote: > Initialise all elements of ar_low with 0 > > for (int i=0; i<n-1; i++) { > for (int j=i+1; j<n-1; j++) { > if (ar[j] <= ar[i]) > ar_low[i]++ ; > } > } > > O(n^2) > > > For O(nlogn), create a BST - O(nlogn) > Traverse the tree, counting children on the left side of each node and > putting it ar_low - 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.
