// sort ar[]
and store in temp[] -- o(nlogn)
for(i=0 to n-1)
{
//search position of ar[i] in temp[] binary search --o(logn)
ar_low[i] = pos-1;
delete temp[pos];
}
in binary search increase pos until next element is same .
On Tue, Jul 13, 2010 at 4:09 PM, Amir hossein Shahriari <
[email protected]> wrote:
> make a balanced bst which also has the size of subtree at each node
> start from ar[n-1] and insert each element and see what is it's rank in BST
> for finding the rank when inserting each time you pick the right subtree
> add size of left subtree to rank
> O(nlogn)
>
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