I just bumped on to this question today and was trying for a solution that is better than O(N) but could not come up with one.
Searched through SO but couldn\'t find
Yet,there is another logic...
(Even numbers are stored in even address)
First check whether the search element is odd or even
If the search element is"even",then perform search only for even adress(Create loop increment to skip odd address)
For example:
You can divide the element and search in parallel or Use "pivot key" to sort to rest 50 elements or any other search method