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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