Peeking Iterator Python Solution with Tests

lcpy gen -n 284   # by problem number
lcpy gen -s peeking_iterator   # by problem name

Input
["PeekingIterator", "next", "peek", "next", "next", "hasNext"]
[[[1, 2, 3]], [], [], [], [], []]
Output
[null, 1, 2, 2, 3, false]

Explanation
PeekingIterator peekingIterator = new PeekingIterator([1, 2, 3]); // [1,2,3]
peekingIterator.next();    // return 1, the pointer moves to the next element [1,2,3].
peekingIterator.peek();    // return 2, the pointer does not move [1,2,3].
peekingIterator.next();    // return 2, the pointer moves to the next element [1,2,3]
peekingIterator.next();    // return 3, the pointer moves to the next element [1,2,3]
peekingIterator.hasNext(); // return False

class Iterator:
    def __init__(self, nums: list[int]) -> None:
        self._nums: list[int] = list(nums)
        self._index: int = 0

    def next(self) -> int:
        value = self._nums[self._index]
        self._index += 1
        return value

    def has_next(self) -> bool:
        return self._index < len(self._nums)


class PeekingIterator(Iterator):
    # Time: O(1) per call
    # Space: O(1)
    def __init__(self, iterator: Iterator) -> None:
        self._iterator = iterator
        self._peeked = 0
        self._has_peeked = False

    def peek(self) -> int:
        if not self._has_peeked:
            self._peeked = self._iterator.next()
            self._has_peeked = True
        return self._peeked

    def next(self) -> int:
        if self._has_peeked:
            self._has_peeked = False
            return self._peeked
        return self._iterator.next()

    def has_next(self) -> bool:
        return self._has_peeked or self._iterator.has_next()