Merge One List Between Two Positions

list1 and list2 contain linked-list node values in traversal order. Remove the inclusive range of nodes from index a through index b in list1, and splice all nodes of list2 in their place.
Return the resulting node values.

Function
mergeInBetween(list1: int[], a: int, b: int, list2: int[]) → int[]

Examples
Example 1
list1 = [0,1,2,3,4,5]
a = 3
b = 4
list2 = [1000000,1000001,1000002]
return = [0,1,2,1000000,1000001,1000002,5]
Values at indices 3 and 4 are replaced by the second list.
Example 2
list1 = [1,2,3]
a = 0
b = 2
list2 = [9]
return = [9]
Replacing the full first list leaves only the inserted list.

Constraints
1 <= list1.length <= 100000.
0 <= a <= b < list1.length.
0 <= list2.length <= 100000.

The trick is that it's just array slicing. Take list1 from 0 up to a (exclusive), append all of list2, then append list1 from b+1 to the end. Three pieces, one pass, O(n + m) time. The pitfall is off-by-one errors. The range is inclusive, so the tail starts at b+1, not b. Check your slices against Example 2, where a is 0 and b is the last index, so both the head and tail come out empty and only list2 remains. Also handle an empty list2, which should simply delete the range. Don't simulate pointers or shift elements one by one in a loop, since with lists of 100000 that can turn quadratic. If the assessment goes sideways and you freeze on the boundaries, StealthCoder is the hedge that reads the problem and hands you the slice logic live.

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