@sunny : this will work :)

On Wed, Dec 14, 2011 at 4:08 PM, sunny agrawal <sunny816.i...@gmail.com>wrote:

> i am Considering given heap as min heap
> Sol -  because heap has property that root will has lower value than all
> the elements in its left sub tree and right sub tree
> so in main we will call a function passing root and value k and x
> if at any time root is greater than x and k > 0 that means rest of the
> elements are greater than x so kth is also greater than x
>
> else make recursive calls for both of its child as soon as k hits zero in
> any recursive call we know that there are k elements less than x.
>
> i think in worst case 2k comparisons will be there hence O(k)
>
> On Wed, Dec 14, 2011 at 12:24 PM, atul anand <atul.87fri...@gmail.com>wrote:
>
>> yup rite...it should be O(k log n ) not O(n log n).
>>
>>
>> On Wed, Dec 14, 2011 at 11:44 AM, Dave <dave_and_da...@juno.com> wrote:
>>
>>> @Atul: The initial heap is given. You have to maintain the heap
>>> property as k elements are removed, which is O(k log n). This does not
>>> satisfy the original request for an algorithm that is "O(k) in the
>>> worst-case, independent of the size of the heap."
>>>
>>> Dave
>>>
>>> On Dec 13, 10:31 pm, atul anand <atul.87fri...@gmail.com> wrote:
>>> > @gaurav : you need to first build heap and then maintain heap property
>>> ever
>>> > time you remove element.so this would take O(n logn ).
>>> >
>>> >
>>> >
>>> > On Wed, Dec 14, 2011 at 1:38 AM, Gaurav Kumar <gkuma...@gmail.com>
>>> wrote:
>>> > > Why can't we keep removing the minimum element each time and compare
>>> it
>>> > > with x? This should take O(k) time since in a Min heap, the minimum
>>> element
>>> > > can be removed in O(1) time? Am I missing something?
>>> >
>>> > > On Tue, Dec 13, 2011 at 8:43 AM, atul anand <atul.87fri...@gmail.com
>>> >wrote:
>>> >
>>> > >> O(k) in the worst-case , then i guess it would better to use
>>> > >> median-of median algo to find element at rank k. and comparing with
>>> x.
>>> >
>>> > >> or
>>> > >> we can us hashtable to solve this.
>>> >
>>> > >> On Tue, Dec 13, 2011 at 3:23 PM, Ankur Garg <ankurga...@gmail.com>
>>> wrote:
>>> >
>>> > >>> Given an array-based heap on n elements and a real number x,
>>> efficiently
>>> > >>> determine whether the kth smallest element in the heap is greater
>>> than or
>>> > >>> equal to x. Your algorithm should be O(k) in the worst-case,
>>> independent of
>>> > >>> the size of the heap.
>>> >
>>> > >>> This question was also asked in Amazon
>>> >
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>
>
>
> --
> Sunny Aggrawal
> B.Tech. V year,CSI
> Indian Institute Of Technology,Roorkee
>
>
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