Our last problem had a sorted dataset as the input. Sure, it was rotated, but **it started out sorted.**

A sorted list gives us ways to leverage that ordering. After all, someone went through a lot of work putting that data in order!

*Binary search* is an algorithm which finds a target value in sorted datasets in `O(logN)`

time using knowledge of the dataset to guide the search.

We check the middle value. If it matches our target, we’ve found the value. If the middle value is **greater than** our target, we can safely ignore everything with a larger index because those values will **also be greater**. This cuts the problem in half.

Our target is the element with a unique property: the **values at the previous and following index both have a greater value.**

This is only true at the index where the rotation “finished” because it marks the beginning of a sorted list.

rotated = ['c', 'd', 'e', 'f', 'a', 'b'] # rotation index is 4 'a' < 'b' # True 'a' < 'f' # True

Use a modified binary search algorithm to improve our solution’s time efficiency from `O(N)`

to `O(logN)`

.

### Instructions

**1.**

Write a function `count_rotations()`

which has one parameter `rotated_list`

.

`count_rotations()`

should return the index where the sorted list would begin.

**Your solution should make use of binary search for a O(logN) time complexity!**