Searching in a sorted and rotated array

This can be done in O(logN) using a slightly modified binary search.

The interesting property of a sorted + rotated array is that when you divide it into two halves, atleast one of the two halves will always be sorted.

Let input array arr = [4,5,6,7,8,9,1,2,3]
number of elements  = 9
mid index = (0+8)/2 = 4

[4,5,6,7,8,9,1,2,3]
         ^
 left   mid  right

as seem right sub-array is not sorted while left sub-array is sorted.

If mid happens to be the point of rotation them both left and right sub-arrays will be sorted.

[6,7,8,9,1,2,3,4,5]
         ^

But in any case one half(sub-array) must be sorted.

We can easily know which half is sorted by comparing start and end element of each half.

Once we find which half is sorted we can see if the key is present in that half – simple comparison with the extremes.

If the key is present in that half we recursively call the function on that half
else we recursively call our search on the other half.

We are discarding one half of the array in each call which makes this algorithm O(logN).

Pseudo code:

function search( arr[], key, low, high)

        mid = (low + high) / 2

        // key not present
        if(low > high)
                return -1

        // key found
        if(arr[mid] == key)
                return mid

        // if left half is sorted.
        if(arr[low] <= arr[mid])

                // if key is present in left half.
                if (arr[low] <= key && arr[mid] >= key) 
                        return search(arr,key,low,mid-1)

                // if key is not present in left half..search right half.
                else                 
                        return search(arr,key,mid+1,high)
                end-if

        // if right half is sorted. 
        else    
                // if key is present in right half.
                if(arr[mid] <= key && arr[high] >= key) 
                        return search(arr,key,mid+1,high)

                // if key is not present in right half..search in left half.
                else
                        return search(arr,key,low,mid-1)
                end-if
        end-if  

end-function

The key here is that one sub-array will always be sorted, using which we can discard one half of the array.