if we do not know k, how do we find it?

inthe series : 17 15 13 11 9 1 3 5 7

we can take k at element 1 also as the descending series can be 17 15 13 11
9 1 and ascending 3 5 7
OR
descending 17 15 13 11 9 and ascending 1 3 5 7
isn't there any relationship in kth and (k+1)th element that helps us find
out k correctly

On Thu, Aug 4, 2011 at 7:37 AM, Ankur Khurana <[email protected]>wrote:

> For question two ,
> try this.
>
> see te element at arr[0] . and supose you have to find k . if arr[0]==k ,
> then yes we found the element else see the diff betweem arr[0] and k. that
> will be minimum amount of steps needed to convert a[0] to k(let abs(a[0]-k)
> = p). then repeat the procedure again at the new element arr[p] untill you
> find the number of reach end of array......
>
>
> On Thu, Aug 4, 2011 at 6:59 AM, Dave <[email protected]> wrote:
>
>> @Amit: If k is not known, you can find it with another binary search.
>>
>> Dave
>>
>> On Aug 3, 3:02 pm, amit karmakar <[email protected]> wrote:
>> > I think for question 1, the value of k is not provided, right?
>> >
>> > On Aug 4, 12:53 am, Ankur Garg <[email protected]> wrote:
>> >
>> >
>> >
>> > > Dave's solution looks gud to me :)
>> >
>> > > On Wed, Aug 3, 2011 at 3:52 PM, Ankur Garg <[email protected]>
>> 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 <[email protected]>
>> 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 <[email protected]> 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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>> > - Show quoted text -
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>
>
> --
> Ankur Khurana
> Computer Science
> Netaji Subhas Institute Of Technology
> Delhi.
>
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-- 
Tushar Bindal
Computer Engineering
Delhi College of Engineering
Mob: +919818442705
E-Mail : [email protected]
Website: www.jugadengg.com

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