On 11 April 2010 10:46, Rohit Saraf <rohit.kumar.sa...@gmail.com> wrote:

> Construct a binary tree from the data (maintain the size of subtree under
> each node).
> Do rotations till the left subtree does not have size k. Rotation is a
> constant time operation.
> Please prove the correctness of your algorithm with the time complexity
> --------------------------------------------------
> Rohit Saraf
> Second Year Undergraduate,
> Dept. of Computer Science and Engineering
> IIT Bombay
> http://www.cse.iitb.ac.in/~rohitfeb14<http://www.cse.iitb.ac.in/%7Erohitfeb14>
>
>
>
> On Mon, Mar 29, 2010 at 11:15 AM, blackDiamond <patidarc...@gmail.com>wrote:
>
>> nice solution appreciate it. but your algorithm is wasting time in finding
>> all the element...
>> instead of that just find boundary line kth element which can help as in
>> finding element greater that kth and element small than kth and that soluton
>> can be done in O(N)
>>
>>
>> On Sun, Mar 28, 2010 at 10:02 PM, CHERUVU JAANU REDDY <
>> jaanu.cher...@gmail.com> wrote:
>>
>>>
>>> 1) Construct max heap by taking first k elements in an array
>>> 2) if k+1 element less than root of max heap
>>>        a) Delete root of max heap
>>>        b) Insert k+1 element in max heap and apply heapify method
>>> 3) else skip the  element
>>> 4) apply above procedure for all n elements in an array
>>>
>>> At last you will get k smallest elements and root is kth smallest element
>>> in the array
>>>
>>> this is O(nlogk)
>>>
>>>
>>>
>>> ----------------------------------------
>>> CHERUVU JAANU REDDY
>>> M.Tech in CSIS
>>>
>>>
>>> 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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>>>>>
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-- 
Thanks & Regards,
Priyanka Chatterjee
Third Year Undergraduate Student,
Computer Science & Engineering,
National Institute Of Technology,Durgapur
India
http://priyanka-nit.blogspot.com/

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