even if u maintain the size , u never know whats the size of child
left-subtree and right subtree to evaluate.

can you explain the first appraoch ?


On Wed, Dec 9, 2009 at 9:24 PM, Rohit Saraf <[email protected]>wrote:

> u can maintain the size..
> and if you don't want that , at least memoize it.. it won't be O(n^2) then.
>
> Rohit Saraf
> Sophomore
> Computer Science and Engineering
> IIT Bombay
>
>
>
> On Wed, Dec 9, 2009 at 9:12 PM, chitta koushik <[email protected]
> > wrote:
>
>> Though i couldn't get both ur approaches clearly , guess 2nd approach shud
>> take O(n^2)  since you go to each node and find the size of left subtree .
>>
>> Or i shud be missing something. my bad can you explain your approaches
>> clearly.
>>
>> On Wed, Dec 9, 2009 at 8:20 PM, Rohit Saraf 
>> <[email protected]>wrote:
>>
>>> do it iteratively either by:
>>>
>>> 1) If size of left tree is less than k, rotate the tree left. and so on
>>> till .....single while loop required for this.
>>> or
>>> 2) Start from head, if k is more than size of left-tree, go to left and
>>> continue searching.. other wise go right and search for k-size(left)-1 in
>>> right tree. All this can be implemented in a single while loop.
>>>
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