you have to find the kth element of your list. which will take O(n)
and remaining element you can get in the other traversal
eg: 3,5,4,35,599,34
you have to find the 3 min element from the lsit
so finding 3rd element we will get 5
on next thing is traverse the list and take all the element less than 5,
eg:3,5,4
is this clear your doubt ...? or i have gone on the other way...

On Sun, Mar 28, 2010 at 8:41 PM, abhijith reddy <abhijith200...@gmail.com>wrote:

> Can any one tell how to do this when there are 'm' queries like "query i j
> k" find the kth largest element in between indices i->j in an array.
> When m is large even an O(n) algorithm would be slow.
> I thinking that each query could be answered in O(sqrt(n)) time
> So any suggestions ?
>
> Thanks
>
>
> On Sun, Mar 28, 2010 at 7:57 PM, blackDiamond <patidarc...@gmail.com>wrote:
>
>> there are better solution of O(n) are posted in the thread.......[?].
>> using order statices ....
>>
>>
>> On Sun, Mar 28, 2010 at 6:49 PM, Mukesh Kumar thakur <
>> mukeshraj8...@gmail.com> wrote:
>>
>>> Create a temp array temp[0..k-1] of size k.
>>> 2) Traverse the array arr[k..n-1]. While traversing, keep updating the
>>> smallest element of temp[]
>>> 3) Return the smallest of temp[]
>>> Time Complexity: O((n-k)*k).
>>>
>>>
>>> try it ..............for this problem[?]
>>>
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