We can also build a Balanced BST for the window elements and on each shift 
we need to have a delete operation and add oeration.
O(n logk)

Also we can add the Shashank algo 
where we check if the newly added element is greater than the current BST's 
max element.
So we can discard the current BSt and start a new BST

Thanks
Ramindar Singh

On Friday, 2 September 2011 09:04:26 UTC+5:30, Anup wrote:
>
> Given an unsorted Array A and any integer k where k <= size of A
>
> Print the maximum of each sub-array of size k of A.
>
> eg:  A = [ 3, 5, 1, 9, 0, 4, -1, 7 ]       k = 4
> Max: 9 9 9 9 7
>
> -- 
> Anup Ghatage
>

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