Leetcode #2353: Design a Food Rating System
In this guide, we solve Leetcode #2353 Design a Food Rating System in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
Design a food rating system that can do the following: Modify the rating of a food item listed in the system. Return the highest-rated food item for a type of cuisine in the system.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Design, Array, Hash Table, String, Ordered Set, Heap (Priority Queue)
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
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".
Python Solution
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)
Complexity
The time complexity is O(n). The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.