Design a Food Rating System Python Solution

lcpy gen -n 2353   # by problem number
lcpy gen -s design_a_food_rating_system   # by problem name

Input
["FoodRatings", "highestRated", "highestRated", "changeRating", "highestRated", "changeRating", "highestRated"]
[[["kimchi", "miso", "sushi", "moussaka", "ramen", "bulgogi"], ["korean", "japanese", "japanese", "greek", "japanese", "korean"], [9, 12, 8, 15, 14, 7]], ["korean"], ["japanese"], ["sushi", 16], ["japanese"], ["ramen", 16], ["japanese"]]
Output
[null, "kimchi", "ramen", null, "sushi", null, "ramen"]

Explanation
foodRatings.highestRated("korean"); // return "kimchi"
// "kimchi" is the highest rated korean food with a rating of 9.
foodRatings.highestRated("japanese"); // return "ramen"
// "ramen" is the highest rated japanese food with a rating of 14.
foodRatings.changeRating("sushi", 16); // "sushi" now has a rating of 16.
foodRatings.highestRated("japanese"); // return "sushi"
foodRatings.changeRating("ramen", 16); // "ramen" now has a rating of 16.
foodRatings.highestRated("japanese"); // return "ramen"
// Both "sushi" and "ramen" have a rating of 16.
// However, "ramen" is lexicographically smaller than "sushi".

import heapq


class FoodRatings:
    # Time: init O(n), change_rating O(log n), highest_rated amortized O(log n)
    # Space: O(n) for the rating/cuisine maps and one lazy heap per cuisine
    def __init__(self, foods: list[str], cuisines: list[str], ratings: list[int]) -> None:
        self.rating: dict[str, int] = dict(zip(foods, ratings, strict=True))
        self.cuisine = dict(zip(foods, cuisines, strict=True))
        self.heaps: dict[str, list[tuple[int, str]]] = {}
        for food, cuisine, rating in zip(foods, cuisines, ratings, strict=True):
            self.heaps.setdefault(cuisine, []).append((-rating, food))
        for heap in self.heaps.values():
            heapq.heapify(heap)

    # Time: O(log n)
    # Space: O(1) amortized (each pushed entry is popped at most once)
    def change_rating(self, food: str, new_rating: int) -> None:
        self.rating[food] = new_rating
        # The old entry for this food is left behind as stale; highest_rated
        # discards entries whose rating no longer matches the current one.
        heapq.heappush(self.heaps[self.cuisine[food]], (-new_rating, food))

    # Time: O(log n) amortized
    # Space: O(1)
    def highest_rated(self, cuisine: str) -> str:
        heap = self.heaps[cuisine]
        while True:
            neg_rating, food = heap[0]
            if -neg_rating == self.rating[food]:
                return food
            heapq.heappop(heap)