I guess you want the following juggling algorithm
http://www.cs.bell-labs.com/cm/cs/pearls/s02b.pdf
On Fri, Jul 2, 2010 at 11:16 PM, Dave <dave_and_da...@juno.com> wrote:
> @Jalaj. The original poster said, "P.S---do not give block reversal
> method for array rotation ...."
>
> Dave
>
> On Jul 2, 10:54 am, jalaj jaiswal <jalaj.jaiswa...@gmail.com> wrote:
> > reverse full array first
> > then, reverse last k elemnts and initial n-k elements seperately
> > this will do
> > On Fri, Jul 2, 2010 at 8:34 PM, Ratnesh Thakur <
> ratneshthaku...@gmail.com>wrote:
> >
> >
> >
> >
> >
> > > correction..
> > > a[j]=a[j-1] instead of a[i]=a[i-1]
> >
> > > On Fri, Jul 2, 2010 at 7:30 PM, Ratnesh Thakur <
> ratneshthaku...@gmail.com>wrote:
> >
> > >> i think this should work.
> >
> > >> for(i=1;i<=k;i++)
> > >> {
> > >> var=a[n-1]
> > >> for(j=n-1;j>=1;j--)
> > >> a[i]=a[i-1]
> > >> a[0]=var
> >
> > >> }
> >
> > >> On Fri, Jul 2, 2010 at 5:36 PM, Saurabh Ahuja <nsit.saur...@gmail.com
> >wrote:
> >
> > >>> a[0] = a[2]
> > >>> a[1] = a[3]
> > >>> a[2] = a[4]
> >
> > >>> a[0] and a[1] has been changed
> > >>> a[3] = a[0]
> > >>> a[4] = a[1]
> >
> > >>> so this solution would not work.
> >
> > >>> On Fri, Jul 2, 2010 at 5:14 PM, Akash Gangil <akashg1...@gmail.com
> >wrote:
> >
> > >>>> wouldn't this work:
> >
> > >>>> for i in range(0,len)
> > >>>> a[i] = a[(i+2)%5];
> >
> > >>>> where len is the length of array
> >
> > >>>> On Sat, Jun 26, 2010 at 3:37 PM, sharad kumar <
> sharad20073...@gmail.com
> > >>>> > wrote:
> >
> > >>>>> i have to right rotate an array by k positions
> > >>>>> 1 2 3 4 5 for k=2 o/p shud be
> > >>>>> 3 4 5 1 2
> >
> > >>>>> P.S---do not give block reversal method for array rotation and soln
> > >>>>> must be inplace.....plzz write ur logic also along with d code
> >
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> > >>>> Akash Gangil
> >
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> > With Regards,
> > Jalaj Jaiswal
> > +919026283397
> > B.TECH IT
> > IIIT ALLAHABAD- Hide quoted text -
> >
> > - Show quoted text -
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