oh common shashank...its not that easy u told.....
just do coding....then u will find the error in ur logic

On Fri, Aug 12, 2011 at 6:25 PM, WgpShashank <shashank7andr...@gmail.com>wrote:

> Hi Guys Here is the algorithm for doing same in O(logn)
>
> Hint :  Find the pivot point, divide the array in two sub-arrays and call
> binary search. Thats It :)
>
> Algorithm
> 1. A Pivot Point is point around which array is rotated so 1st we need to
> find out its location/index. lets say it pivot.
>       a. to find pivot point we use same idea of binary search  & check if
> arr[mid]>a[mid+1] if true then return pivot=mid.
>       b  else if arr[low]<a[mid] search for pivot in left
>       c. else search for pivot in right half of array.
>      Time Complexity O(logn)
>
> 2.  Once we have found index of pivot point check if desired element is
> same value at pivot index or not e.g.
>     a. ( if a[pivot]==value we are searching for ) the simply return true
> or 1 .
>           Now call binary search for one of the two sub-arrays.
>          (ab) *If *element is greater than 0th element then  search in
> left array
>          (ac) *Else* Search in right array  *
>          (ad) If *element is found in selected sub-array then return
> index
>                     *Else *return -1.
>
>          Again Time Complexity O(logn)
>
> Hope You Can Make it in Running Code , Do Noify me if You Need More
> Explanation or if missed Something??
>
> *Regards
> Shashank Mani "Computer Science Is Awesome So Why I Write Code"
> Computer Science
> Birla institute of Technology Mesra
> *
>
>
>
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-- 
**Regards
SAGAR PAREEK
COMPUTER SCIENCE AND ENGINEERING
NIT ALLAHABAD

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