I guess it is a dynamic programming problem.
On Wed, Jul 14, 2010 at 11:34 AM, Anand <anandut2...@gmail.com> wrote:
> One approach will be while creating a BST and also store position of the
> element in the original array. While constructing ar_low check for two
> condition 1. element should be less than the given element and also position
> of element should greater than the given element.
>
>
> On Tue, Jul 13, 2010 at 4:01 AM, Vipul Agrawal <vipul.itbh...@gmail.com>wrote:
>
>> // 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 <
>> amir.hossein.shahri...@gmail.com> 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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Thanks,
Chakravarthi.
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