How do you solve Design a Food Rating System on LeetCode?

Design a Food Rating System (LeetCode #2353) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(log n) for changeRating and O(1) for highestRated. time complexity and O(n) space complexity. Key insight: Using a hash map allows for O(1) access to food ratings.

What is the time complexity of Design a Food Rating System?

The optimal solution for Design a Food Rating System runs in O(log n) for changeRating and O(1) for highestRated. time and uses O(n) space. The complexity is efficient because we only need to maintain a heap for each cuisine, allowing us to quickly access and update the highest-rated food.

What is the brute force solution for Design a Food Rating System?

The brute force approach for Design a Food Rating System is Brute Force, with O(n²) time complexity and O(1) space complexity. In this approach, we will iterate through all food items to find the highest-rated food for a given cuisine. This is straightforward but inefficient as it checks every food item each time we need the highest rating. The optimal Optimal Solution approach improves this to O(log n) for changeRating and O(1) for highestRated..

What are the interview tips for Design a Food Rating System?

Explain your thought process clearly while coding. Discuss potential edge cases, like ties or invalid food names. Practice coding the heap operations beforehand.

What are common mistakes when solving Design a Food Rating System?

Not updating the heap properly after changing ratings. Forgetting to handle ties in ratings correctly.

#2353

Design a Food Rating System

Medium

ArrayHash TableStringDesignHeap (Priority Queue)Ordered SetHash MapPriority Queue

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(log n) for changeRating and O(1) for highestRated.
Space	O(1)	O(n)

💡

Intuition

Time O(log n) for changeRating and O(1) for highestRated.Space O(n)

In this approach, we will use a combination of a hash map and a max heap (priority queue) to efficiently manage food ratings and quickly retrieve the highest-rated food for a given cuisine.

⚙️

Algorithm

4 steps

1Step 1: Use a hash map to map each food to its current rating and cuisine.
2Step 2: Use a max heap (priority queue) to maintain the highest-rated food items for each cuisine.
3Step 3: For changeRating, update the rating in the hash map and reinsert the food into the heap to maintain order.
4Step 4: For highestRated, simply retrieve the top of the heap for the specified cuisine.

solution.py26 lines

1import heapq
2
3class FoodRatings:
4    def __init__(self, foods, cuisines, ratings):
5        self.food_map = {}
6        self.cuisine_map = {}
7        self.cuisine_heap = {}
8
9        for food, cuisine, rating in zip(foods, cuisines, ratings):
10            self.food_map[food] = rating
11            self.cuisine_map[food] = cuisine
12            if cuisine not in self.cuisine_heap:
13                self.cuisine_heap[cuisine] = []
14            heapq.heappush(self.cuisine_heap[cuisine], (-rating, food))
15
16    def changeRating(self, food, newRating):
17        cuisine = self.cuisine_map[food]
18        self.food_map[food] = newRating
19        heapq.heappush(self.cuisine_heap[cuisine], (-newRating, food))
20
21    def highestRated(self, cuisine):
22        while self.cuisine_heap[cuisine]:
23            rating, food = self.cuisine_heap[cuisine][0]
24            if -rating == self.food_map[food]:
25                return food
26            heapq.heappop(self.cuisine_heap[cuisine])

ℹ

Complexity note: The complexity is efficient because we only need to maintain a heap for each cuisine, allowing us to quickly access and update the highest-rated food.

1Using a hash map allows for O(1) access to food ratings.
2A max heap efficiently keeps track of the highest-rated food.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.