Merge Two Sorted Arrays

Given two integer arrays first and second, each sorted in non-decreasing order, return a new array containing every value from both inputs in non-decreasing order.
Retain every occurrence of a value, including duplicates within one input or across both inputs.

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

Examples
Example 1
first = [1,2,4]
second = [1,3,4]
return = [1,1,2,3,4,4]
The two values equal to 1 and the two values equal to 4 are all retained in sorted order.
Example 2
first = []
second = [2,2]
return = [2,2]
The first array is empty, so the result contains both values from second.
Example 3
first = [-5,-1,0]
second = [-3,2]
return = [-5,-3,-1,0,2]
Values from the two arrays alternate in the merged ordering.

Constraints
0 <= first.length <= 200000
0 <= second.length <= 200000
Every value is a signed 32-bit integer.
Both input arrays are sorted in non-decreasing order.

The trick is two pointers, one per array. Compare first[i] and second[j], push the smaller, advance that pointer. Use <= when comparing so ties take from either side without losing a value. Duplicates are retained automatically because you never skip anything. The pitfall is the tail. When one array runs out, you must copy the rest of the other. Forgetting that, or only handling it for one side, fails Example 2 and any case where one array is exhausted early. Don't concatenate and sort. That's O(n log n) when O(n + m) is expected, and with 200000 elements each it's wasteful. Preallocate the result at size n + m. Negative values and 32-bit extremes don't matter since you only compare, never subtract. If you freeze on the OA, StealthCoder is the hedge that hands you the loop, but the logic is about ten lines.

If you see this problem in your OA tomorrow, the play is to recognize the pattern in 30 seconds. StealthCoder buys you that recognition.