@souravsain:
Given an array of n elements and an integer k where k<n. Elemnts
> > {a[0].....a[k] and a[k+1].....a[n] are already sorted.
As per the question a{0} to a{k} is sorted and a{K+1} to a{n} is sorted
so we look at the sequence in a{0} to a{k} and {n} to a{k+1} it makes a
bitonic sequence.
and if we apply bitonic merge on it, it gives a final sorted sequence.
On Sat, Jul 3, 2010 at 12:23 AM, souravsain <souravs...@gmail.com> wrote:
> @Anand
>
> Please explain how you concluded that the array will first
> continuously increase and then continuously decrease? Why can it not
> be 2 continuous increase like [1,2,3,4,5,3,4,8] where [1,2,3,4,5] and
> [3,4,8] are a[1] to a[k] and a[k+1] to a[N] respectively? Whill your
> method work still?
>
> @Ankur, Correct me if my interpretation of the question is wrong.
>
> Sourav
>
> On Jul 3, 1:32 am, Anand <anandut2...@gmail.com> wrote:
> > This is an example of bitonic sequence if we reverse the bottom half of
> the
> > array.
> > Sequence is called Bitonics if the sequence of number first
> > increases(ascending order) and then decrease(descending order).
> >
> > 1. We need to reverse the bottom half the array to make it bitonic.
> > 2. Appy Bitonic Merge to get the final sorted array.: Complexity.O(n)
> >
> > In the below code, I have implemented sorting n/w to sort any kind of
> array
> > but for bitonic sequence we only bitonic merge function call which take
> > O(n).
> > Refer section Sorting network from Corman for more details
> >
> > http://codepad.org/ZhYEBqMw
> >
> > On Fri, Jul 2, 2010 at 11:30 AM, ANKUR BHARDWAJ <ankibha...@gmail.com
> >wrote:
> >
> > > Given an array of n elements and an integer k where k<n. Elemnts
> > > {a[0].....a[k] and a[k+1].....a[n] are already sorted. Give an
> > > algorithm to sort in O(n) time and O(1) space.
> >
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