You don't need any data structure.
Since the window moves by one element each, you can simply count 10
such moves and compare each new element with currMax.
If it's greater, overwrite currMax. After 10 moves you have your max
for that 10 element window, rinse and repeat.


On Sep 17, 1:31 am, Navin Naidu <navinmna...@gmail.com> wrote:
> Use Max-Heap, add first ten elements, the root element will be max,
>
> Next Iteration, remove a[0] and add a[10], max-heapify.
>
>
>
>
>
>
>
>
>
> On Fri, Sep 17, 2010 at 1:48 PM, Tech Id <tech.login....@gmail.com> wrote:
> > You have an array of 100 integers.
> > There is a window of 10 elements which starts from 0th element and
> > moves by 1 element till the 90th element.
> > We need to print the maximum element for all the positions of the
> > window.
>
> > Hint:
> > For the first position of the window, you have to find the maximum as
> > done normally for any array.
> > After that, when the window moves by 1 element, you just have to
> > consider one more element and forget the very first element to find
> > the new maximum. Thus, after the first max, it should take much less
> > time to find the max for all the other window positions.
>
> > --
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> --
> Thanks & Regards,
>
> - NMN

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