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.



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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