Make a Sorted Array Unique 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
makeUniqueInPlace(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 two pointers. Keep a write index starting at 1 and scan with a read index from 1. Whenever nums[read] differs from nums[read-1], copy it to nums[write] and bump write. Because the array is sorted, duplicates are always adjacent, so comparing neighbors is enough. At the end, return the first write elements. The common pitfall is deleting elements mid-loop, which shifts everything and turns O(n) into O(n^2) at n up to 10^5. The second pitfall is forgetting the empty array: guard for length 0 and return [] before touching index 0. Also compare against the previous read value or the last written value, either works, but don't mix them up. In a shared doc with no test runner, trace Example 2 by hand before you finish. If you freeze live, StealthCoder is the hedge that gives you the pointer logic fast.

If this hits your live OA and you blank, StealthCoder solves it in seconds, invisible to the proctor.