Merge Two Sorted Collections Without Duplicates

You are given two integer arrays first and second, each sorted in nondecreasing order.
Return one nondecreasing integer array that contains every value appearing in either input exactly once. Duplicate occurrences within one input or across both inputs must contribute only one value to the result.

Function
mergeSortedUnique(first: int[], second: int[]) → int[]

Examples
Example 1
first = [1,2,2,4]
second = [2,3,4,4,5]
return = [1,2,3,4,5]
The value 2 appears in both inputs and 4 appears more than once, but each is included only once in the merged result.
Example 2
first = []
second = [-3,-3,0,7]
return = [-3,0,7]
The first input is empty. Removing the repeated -3 from the second input leaves the complete result.
Example 3
first = [1,1,1]
second = [1,1]
return = [1]
All elements have the same value, so the unique union contains one element.

Constraints
0 <= first.length, second.length <= 100000.
-10^9 <= first[i], second[i] <= 10^9.
Both input arrays are sorted in nondecreasing order.

The trick: both inputs are already sorted, so don't reach for a set and a re-sort. That works but costs O(n log n) and ignores the gift in the constraints. Use two pointers, i and j. At each step, pick the smaller of first[i] and second[j], and append it only if the result is empty or differs from the last appended value. Advance the pointer you took from. When values are equal, either one works, since the last-value check drops the duplicate. Then drain whichever array has leftovers using the same check. That's O(n+m) time and O(1) extra beyond the output. Common pitfalls: comparing against the input neighbor instead of the last output value, forgetting to drain the tail, and crashing on an empty array. Check [] with [-3,-3,0,7] and [1,1,1] with [1,1] before submitting. If you freeze during the live OA, StealthCoder can hand you the clean two-pointer version.

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