are yaar... i meant BST... i thought that was obvious !
sry if i confused you....


--------------------------------------------------
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
<[email protected]>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 <[email protected]
> > 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 <[email protected]
>> > 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 <[email protected]> wrote:
>>> >
>>> >
>>> > On Sun, Apr 11, 2010 at 9:56 PM, Rohit Saraf <
>>> [email protected]>
>>> > 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
>>> >> <[email protected]> wrote:
>>> >>>
>>> >>>
>>> >>>
>>> >>> On 11 April 2010 10:46, Rohit Saraf <[email protected]>
>>> 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 <
>>> [email protected]>
>>> >>>> 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
>>> >>>>> <[email protected]> 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
>>> >>>>>> <[email protected]> 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 <
>>> [email protected]>
>>> >>>>>>> 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
>>> >>>>>>>> <[email protected]> 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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>>> >
>>> >
>>> >
>>> >
>>> > --
>>> > 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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