For the last problem our suggested solutions had `O(N)`

time and/or `O(N)`

space complexity.

We can’t improve the time complexity. We have to visit each value to determine whether or not it is a duplicate.

We can reduce the space complexity to `O(1)`

with an in-place solution.

We’ll adjust the problem to allow for this space complexity. Now we want to alter a **sorted** input list with **all duplicates moved to the end of the list.**

The return value will be the index of the final unique value.

duplicates = ['a', 'a', 'g', 't', 't', 'x', 'x', 'x'] remove_dups(duplicates) # 3, index of the last unique value: 'x' duplicates # ['a', 'g', 't', 'x' 'a', 'x', 't', 'x']

**Clarifying Questions:**

- Does the ordering of the duplicate(s) matter?
- No! They can be in any order.

### Instructions

**1.**

Write a function `move_duplicates()`

which has one parameter: `dupe_list`

.

The argument to `move_duplicates()`

will always be a sorted list which may have duplicated values.

`move_duplicates()`

should move all duplicate values to the end of the list and maintain sorted order.

The return value is the index of **the last unique value**.

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