@Amit: If k is not known, you can find it with another binary search.

Dave

On Aug 3, 3:02 pm, amit karmakar <amit.codenam...@gmail.com> wrote:
> I think for question 1, the value of k is not provided, right?
>
> On Aug 4, 12:53 am, Ankur Garg <ankurga...@gmail.com> wrote:
>
>
>
> > Dave's solution looks gud to me :)
>
> > On Wed, Aug 3, 2011 at 3:52 PM, Ankur Garg <ankurga...@gmail.com> wrote:
> > > Q1 can be looked as rotated sorted array...check whether the no is less or
> > > greater than kth element ..if greater search using binary search with low 
> > > =0
> > > high k-1 and if less earch in right with low=k+1 high =n;
>
> > > q2) Dont know :(
>
> > > On Wed, Aug 3, 2011 at 3:44 PM, Dave <dave_and_da...@juno.com> wrote:
>
> > >> @Tushar: For problem 1, do a binary search on elements 1 to k, and if
> > >> no hit is found, do a binary search on elements k+1 to n.
>
> > >> For problem 2, suppose that you are searching the given array for the
> > >> number 2. The idea is to take big steps when you are far from the
> > >> target, and small steps when you are close. Start with i = 0. If a[i] !
> > >> = 2, then add abs(a[i]-2) to i and try again. This is because it will
> > >> take at least abs(a[i]-2) steps to get to 2.
>
> > >>  In this case, i = 0 and a[0] = 6, so add 4 to i, getting 4. a[4] = 4,
> > >> so add 2 to i, getting 6. a[6] = 3, so add 1. a[7] = 2.
>
> > >> Dave
>
> > >> On Aug 3, 2:09 pm, TUSHAR <tusharkanta.r...@gmail.com> wrote:
> > >> > 1.   Given an array of n-elements ? the 1st k -elements are in
> > >> > descending order and k+1 to n elements are in
> > >> >       ascending order. give an  efficient algo for searching an
> > >> > element ?
>
> > >> > 2.  Given an array of n-elements ? each element in the array is either
> > >> > same or less by 1 or larger by 1 from the
> > >> >      previous element . give an  efficient algo for searching an
> > >> > element ?
>
> > >> >           e.g :   6 6 6 5 4 4 3 2 3 4 3 4 ............
>
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