Sorry its wrongly posted :(
Please Avoid it.
Well thanks for suggestion @Miroslav.


On Mar 6, 10:19 am, Kapil <navka...@gmail.com> wrote:
> @Prunthaban
> In worst case Miroslav's solution also goes for O(n).All elements are
> duplicate
> But in the case(what i have suggested that i am going to find the
> border and not the element(key)
> So the order has to be O(log n)
>
> On Mar 4, 5:01 pm, Prunthaban <pruntha...@gmail.com> wrote:
>
> > @sharad - Your solution is O(n) Miroslav's solution is O(logn).
> > And what are you doing with the 'j' variable in your solution? Why not
> > simply use a[i] == k then c = i. That is what eventually your solution
> > is doing and it is O(n).
>
> > On Mar 4, 4:32 pm, sharad kumar <aryansmit3...@gmail.com> wrote:
>
> > > pls tell wats difference btwn increasing and non decresing sorted
> > > array.question clearly tells n sorted array.......
>
> > > On Wed, Mar 4, 2009 at 4:55 PM, Miroslav Balaz 
> > > <gpsla...@googlemail.com>wrote:
>
> > > > of course, you cannot assume that array is in ascending order, it is in
> > > > non-decreasing order however not in ascending
> > > > and you should swap order here :(a[mid+1] > key||mid==high)
>
> > > > 2009/3/4 Kapil <navka...@gmail.com>
>
> > > >> just fixing a bug
>
> > > >> what if you write bin_search as this
> > > >> //assumption array is in ascending order
> > > >> binsearch(high, low, key, a)
> > > >> begin
> > > >>  if low > high
> > > >>   return -1
> > > >>  mid = (high+low)/2
> > > >>  if a[mid] = key And (a[mid+1] > key||mid==high)
> > > >>    return mid
> > > >>  if a[mid] <= key
> > > >>   low = mid+1
> > > >>  else
> > > >>   high = mid - 1
> > > >>  return binsearch(high,low,key,a)
> > > >> end
>
> > > >> On Mar 4, 3:46 pm, Kapil <navka...@gmail.com> wrote:
> > > >> > what if you write bin_search as this
> > > >> > //assumption array is in ascending order
> > > >> > binsearch(high, low, key, a)
> > > >> > begin
> > > >> >  if low > high
> > > >> >    return -1
> > > >> >  mid = (high+low)/2
> > > >> >  if a[mid] = key And a[mid+1] > key
> > > >> >    return mid
> > > >> >  if a[mid] <= key
> > > >> >    low = mid+1
> > > >> >  else
> > > >> >    high = mid - 1
> > > >> >  return binsearch(high,low,key,a)
> > > >> > end
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