To solve this problem, implement a system that tracks food ratings for different cuisines. Focus on using efficient data structures like hash tables and priority queues to manage the ratings, and ensure that food items are returned lexicographically when ratings are tied. Use the right structures to handle updates and queries efficiently.
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.
How GhostInterview Helps
- GhostInterview helps you break down this problem into manageable components, guiding you to focus on optimizing data structures.
- By providing step-by-step hints, GhostInterview helps you refine your understanding of priority queues and hash maps.
- GhostInterview also offers guidance on how to handle edge cases and tie-breaking scenarios in this problem.
Topic Pages
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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.
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