I have already given an O(n) solution. See above !

The BST solution is O(nlogn) but is practically more nice. :)

--------------------------------------------------
Rohit Saraf
Second Year Undergraduate,
Dept. of Computer Science and Engineering
IIT Bombay
http://www.cse.iitb.ac.in/~rohitfeb14


On Mon, Apr 12, 2010 at 1:16 PM, Nikhil Agarwal
<nikhil.bhoja...@gmail.com>wrote:

>
>
> On Mon, Apr 12, 2010 at 12:58 PM, Rohit Saraf <rohit.kumar.sa...@gmail.com
> > wrote:
>
>> are yaar... i meant BST... i thought that was obvious !
>> sry if i confused you....
>>
>
> @rohit It's ok.Actually in this group people come up with very different
> and non-common solutions so its risky to take things 'obviously'.
> Rotation algo has a complexity of O(nlogn)[for constructing a BST] +O(n)
> [for rotating n times]=O(nlogn) .
>
> Till now best algo we have is using heaps which give rise to complexity =
> O(n+klogn)
>
> Please pass on algos having better runtime complexity.
>
>>
>>
>> --------------------------------------------------
>> Rohit Saraf
>> Second Year Undergraduate,
>> Dept. of Computer Science and Engineering
>> IIT Bombay
>> http://www.cse.iitb.ac.in/~rohitfeb14
>>
>>
>> On Mon, Apr 12, 2010 at 12:38 PM, Nikhil Agarwal <
>> nikhil.bhoja...@gmail.com> wrote:
>>
>>> Hey rohit.You were referring to Binary tree.Search keyword was
>>> missing.Because rotation makes no sense in binary tree.Please note binary
>>> tree and BST are different.
>>>
>>> On Mon, Apr 12, 2010 at 12:33 PM, Rohit Saraf <
>>> rohit.kumar.sa...@gmail.com> wrote:
>>>
>>>> Read the slides i uploaded. They explain what rotation does in a BST.
>>>>
>>>> Also you might like to refer to Red Black Trees in CLRS.... that chapter
>>>> explains rotations.
>>>>
>>>> --------------------------------------------------
>>>> Rohit Saraf
>>>> Second Year Undergraduate,
>>>> Dept. of Computer Science and Engineering
>>>> IIT Bombay
>>>> http://www.cse.iitb.ac.in/~rohitfeb14
>>>>
>>>>
>>>> On Mon, Apr 12, 2010 at 8:18 AM, Rohit Saraf <
>>>> rohit.kumar.sa...@gmail.com> wrote:
>>>>
>>>>> but still the binary tree solution is of more practical use.i will
>>>>> explain the solution once i reach my comp
>>>>>
>>>>>
>>>>> On 4/11/10, Nikhil Agarwal <nikhil.bhoja...@gmail.com> wrote:
>>>>> >
>>>>> >
>>>>> > On Sun, Apr 11, 2010 at 9:56 PM, Rohit Saraf <
>>>>> rohit.kumar.sa...@gmail.com>
>>>>> > wrote:
>>>>> >>
>>>>> >> Time complexity is O(n log n). But the last solution I gave has
>>>>> O(n).
>>>>> >>
>>>>> >> What did u not understand abt thesolution
>>>>> >
>>>>> >
>>>>> > @Rohit Please explain how that Binary tree solution works.
>>>>> >>
>>>>> >>
>>>>> >> --------------------------------------------------
>>>>> >> Rohit Saraf
>>>>> >> Second Year Undergraduate,
>>>>> >> Dept. of Computer Science and Engineering
>>>>> >> IIT Bombay
>>>>> >> http://www.cse.iitb.ac.in/~rohitfeb14
>>>>> >>
>>>>> >>
>>>>> >> On Sun, Apr 11, 2010 at 11:00 AM, Priyanka Chatterjee
>>>>> >> <dona.1...@gmail.com> wrote:
>>>>> >>>
>>>>> >>>
>>>>> >>>
>>>>> >>> 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
>>>>> >>>>
>>>>> >>>>
>>>>> >>>>
>>>>> >>>> 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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>>>>> >>>>>>>>
>>>>> >>>>>>>>
>>>>> >>>>>>>>
>>>>> >>>>>>>> --
>>>>> >>>>>>>> ~~~~BL/\CK_D!AMOND~~~~~~~~
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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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>>>>> >
>>>>> >
>>>>> >
>>>>> >
>>>>> > --
>>>>> > Thanks & Regards
>>>>> > Nikhil Agarwal
>>>>> > Junior Undergraduate
>>>>> > Computer Science & Engineering,
>>>>> > National Institute Of Technology, Durgapur,India
>>>>> > http://tech-nikk.blogspot.com
>>>>> >
>>>>> >
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>>>>>
>>>>>
>>>>> --
>>>>>
>>>>> --------------------------------------------------
>>>>> Rohit Saraf
>>>>> Second Year Undergraduate,
>>>>> Dept. of Computer Science and Engineering
>>>>> IIT Bombay
>>>>> http://www.cse.iitb.ac.in/~rohitfeb14
>>>>>
>>>>
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>>>
>>>
>>>
>>> --
>>> Thanks & Regards
>>> Nikhil Agarwal
>>> Junior Undergraduate
>>> Computer Science & Engineering,
>>> National Institute Of Technology, Durgapur,India
>>> http://tech-nikk.blogspot.com
>>>
>>>
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>
>
>
> --
> Thanks & Regards
> Nikhil Agarwal
> Junior Undergraduate
> Computer Science & Engineering,
> National Institute Of Technology, Durgapur,India
> http://tech-nikk.blogspot.com
>
>
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