Design a Food Rating System

Design a food rating system that tracks and updates ratings of foods, finding the highest rated items by cuisine.

array hash table string design heap priority queue

Problem Statement

Design a food rating system that supports tracking and updating food ratings by cuisine. Implement the FoodRatings class that allows food items to be rated and queried efficiently. You need to ensure that food ratings are updated dynamically and queries for the highest-rated food by cuisine return results in lexicographical order in case of ties.

The system should allow you to perform the following operations: Add foods, update ratings for foods, and query for the highest rated food in a given cuisine. You will be given a list of foods, their corresponding cuisines, and initial ratings. When querying the highest rated food for a cuisine, return the food with the highest rating. If there is a tie, return the lexicographically smaller food.

Examples

Example 1

Input: See original problem statement.

Output: See original problem statement.

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 foodRatings = new FoodRatings(["kimchi", "miso", "sushi", "moussaka", "ramen", "bulgogi"], ["korean", "japanese", "japanese", "greek", "japanese", "korean"], [9, 12, 8, 15, 14, 7]); 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" // "sushi" is the highest rated japanese food with a rating of 16. 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".

Constraints

1 <= n <= 2 * 104
n == foods.length == cuisines.length == ratings.length
1 <= foods[i].length, cuisines[i].length <= 10
foods[i], cuisines[i] consist of lowercase English letters.
1 <= ratings[i] <= 108
All the strings in foods are distinct.
food will be the name of a food item in the system across all calls to changeRating.
cuisine will be a type of cuisine of at least one food item in the system across all calls to highestRated.
At most 2 * 104 calls in total will be made to changeRating and highestRated.

Solution Approach

Use a Hash Map for Food Data

To efficiently store and update food ratings, use a hash map where the key is the food name and the value is an object that contains the food's rating and cuisine. This allows constant time lookup for food ratings and ensures fast updates to the ratings.

Priority Queue for Highest Rated Food

For each cuisine, maintain a max-heap (priority queue) to store foods and their ratings. This allows efficient retrieval of the highest-rated food. In the case of ties, the lexicographical order of foods will be used to determine the result.

Efficient Rating Updates

When updating a food's rating, you will need to adjust the corresponding entry in the heap for the appropriate cuisine. This requires removing the old rating and re-adding the updated food rating to maintain the heap's properties.

Complexity Analysis

Metric	Value
Time	O((n + m) \log n)
Space	O(n)

The time complexity for each operation is O(log n) due to the heap operations involved in inserting and removing food items. The space complexity is O(n) since we are storing data for each food in a hash map and priority queue.

What Interviewers Usually Probe

Look for efficient management of both hash maps and priority queues.
The ability to handle tie cases by lexicographical order is a key part of this solution.
The candidate should optimize updates to the rating system to ensure the solution can handle large inputs.

Common Pitfalls or Variants

Common pitfalls

Failing to update the heap correctly after changing a food's rating.
Not handling lexicographical order correctly in case of ties when querying the highest rated food.
Using inefficient data structures that result in poor performance when handling large inputs.

Follow-up variants

Extending the system to support multiple cuisines and more complex rating updates.
Adding additional features such as filtering by food category or cuisine.
Handling edge cases such as empty inputs or invalid cuisine names.

FAQ

How do I efficiently handle updates to food ratings in this problem?

Efficiently handle updates by adjusting the rating in the hash map and updating the priority queue with the new rating for the relevant cuisine.

What is the key data structure for solving this problem?

The key data structure is a hash map to store food ratings and a priority queue to track the highest-rated foods by cuisine.

How should I handle ties in food ratings?

In case of a tie, return the food item that is lexicographically smaller, as specified in the problem statement.

Can I optimize the query for the highest-rated food?

Yes, by using a max-heap (priority queue), you can efficiently get the highest-rated food for any cuisine in O(log n) time.

What are the time and space complexities for this problem?

The time complexity for each operation is O(log n) due to heap operations, and the space complexity is O(n) for storing the data.

terminal

Solution

Solution 1: Hash Table + Ordered Set

We can use a hash table $\textit{d}$ to store the foods for each cuisine, where the key is the cuisine and the value is an ordered set. Each element in the ordered set is a tuple $(\textit{rating}, \textit{food})$, sorted by rating in descending order, and if the ratings are the same, sorted by food name in lexicographical order.

class FoodRatings:

    def __init__(self, foods: List[str], cuisines: List[str], ratings: List[int]):
        self.d = defaultdict(SortedList)
        self.g = {}
        for food, cuisine, rating in zip(foods, cuisines, ratings):
            self.d[cuisine].add((-rating, food))
            self.g[food] = (rating, cuisine)

    def changeRating(self, food: str, newRating: int) -> None:
        oldRating, cuisine = self.g[food]
        self.g[food] = (newRating, cuisine)
        self.d[cuisine].remove((-oldRating, food))
        self.d[cuisine].add((-newRating, food))

    def highestRated(self, cuisine: str) -> str:
        return self.d[cuisine][0][1]


# Your FoodRatings object will be instantiated and called as such:
# obj = FoodRatings(foods, cuisines, ratings)
# obj.changeRating(food,newRating)
# param_2 = obj.highestRated(cuisine)

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