Remove Duplicates From Sorted Array In Place

You are given an integer array nums sorted in nondecreasing order. Modify it in place so that each distinct value appears exactly once and the retained values stay in sorted order.
Return an array containing exactly the compacted prefix. Extra capacity after that prefix may be left unchanged.
What the interview report shared
The interviewer asked an easy coding question: remove duplicate elements in a sorted array in place. The candidate solved it in a shared Google Doc.

Function
removeDuplicates(nums: int[]) → int[]

Examples
Example 1
nums = [1,1,2]
return = [1,2]
The sorted values are 1, 1, and 2. After in-place compaction the unique prefix is [1,2].
Example 2
nums = [0,0,1,1,1,2,2,3,3,4]
return = [0,1,2,3,4]
Adjacent duplicates are overwritten so each retained value appears once. The compacted prefix is [0,1,2,3,4].
Example 3
nums = []
return = []
An empty array has no values to retain, so the compacted prefix is empty.

Constraints
0 <= nums.length <= 10^5.
-10^9 <= nums[i] <= 10^9.
nums is sorted in nondecreasing order.

The trick is that sorted means duplicates sit next to each other. Keep a write index starting at 0 (or 1). Walk the array with a read index. Whenever nums[read] differs from the last kept value, write it at the write index and advance. When you finish, the first k slots hold the unique values in order. Return that prefix, like nums[0:k] in Python or a copy of the first k elements. The common pitfalls: comparing against the wrong neighbor, forgetting the empty array case (length 0 is allowed), and returning the count when this version asks for the compacted array. Don't use a set. It breaks the in-place requirement and wastes the sorted guarantee. With n up to 10^5 this runs in O(n) time and O(1) extra space. If you freeze mid-OA, StealthCoder gives you the two-pointer solution in real time so you can confirm your edge cases.

The honest play: practice the pattern, and have StealthCoder ready for the one you didn't see coming.