RLE Iterator Python Solution with Tests

lcpy gen -n 900   # by problem number
lcpy gen -s rle_iterator   # by problem name

Input
["RLEIterator", "next", "next", "next", "next"]
[[[3, 8, 0, 9, 2, 5]], [2], [1], [1], [2]]
Output
[null, 8, 8, 5, -1]

Explanation
RLEIterator rLEIterator = new RLEIterator([3, 8, 0, 9, 2, 5]); // This maps to the sequence [8,8,8,5,5].
rLEIterator.next(2); // exhausts 2 terms of the sequence, returning 8. The remaining sequence is now [8, 5, 5].
rLEIterator.next(1); // exhausts 1 term of the sequence, returning 8. The remaining sequence is now [5, 5].
rLEIterator.next(1); // exhausts 1 term of the sequence, returning 5. The remaining sequence is now [5].
rLEIterator.next(2); // exhausts 2 terms, returning -1. This is because the first term exhausted was 5, but the second term did not exist. Since the last term exhausted does not exist, we return -1.

from bisect import bisect_left


class RLEIterator:
    # Time: O(1) init, O(log k) next where k is the number of runs
    # Space: O(k) for the prefix counts

    def __init__(self, encoding: list[int]) -> None:
        self.prefix: list[int] = []
        self.values = encoding[1::2]
        self.pos: int = 0
        total = 0
        for count in encoding[::2]:
            total += count
            self.prefix.append(total)

    def next(self, n: int) -> int:
        self.pos += n
        idx = bisect_left(self.prefix, self.pos)
        if idx == len(self.prefix):
            return -1
        return self.values[idx]