Flatten a 2D array with remove

Implement a class Vector2D that walks a 2D list of integers row by row, left to right, and lets the caller delete values while walking.

Vector2D(vec: list[list[int]])   # rows may be empty; the list itself may be empty
hasNext() -> bool                # is there another value to return?
next() -> int                    # return the next value and advance
remove() -> None                 # delete the value returned by the most recent next()
getVector() -> list[list[int]]   # the current rows, after all removals so far

Rules:

• next() raises an error when there are no values left.
• remove() may be called at most once per next(). Calling it before the first next(), or twice without a next() in between, raises an error.
• remove() must not change which value next() returns afterwards: iteration continues with the value that followed the removed one.
• getVector() keeps every row in its place, including rows that became empty, and must not expose internal state that the caller could change.
• The iterator must not modify the caller's input lists.
• Up to 10^4 rows and 10^5 values in total; aim for amortized O(1) hasNext() and next().

Example 1

it = Vector2D([[10, 20], [], [30]])
it.next()       -> 10
it.next()       -> 20
it.remove()
it.getVector()  -> [[10], [], [30]]
it.hasNext()    -> True
it.next()       -> 30
it.remove()
it.hasNext()    -> False
it.getVector()  -> [[10], [], []]

Example 2

it = Vector2D([[5], [6, 7]])
it.next()       -> 5
it.hasNext()    -> True
it.remove()          # removes 5, even though hasNext() already looked at row 1
it.getVector()  -> [[], [6, 7]]
it.next()       -> 6