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Design a Food Rating System

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Design a food rating system that can do the following:

  1. Modify the rating of a food item listed in the system.
  2. Return the highest-rated food item for a type of cuisine in the system. If there is a tie, return the lexicographically smaller item.

Implement the FoodRatings class:

  • FoodRatings(String[] foods, String[] cuisines, int[] ratings) Initializes the system. The food items are described by foods, cuisines and ratings, all of which have a length of n.
    • foods[i] is the name of the i<sup>th</sup> food,
    • cuisines[i] is the type of cuisine of the i<sup>th</sup> food, and
    • ratings[i] is the initial rating of the i<sup>th</sup> food.
  • void changeRating(String food, int newRating) Changes the rating of the food item with the name food.
  • String highestRated(String cuisine) Returns the name of the food item that has the highest rating for the given type of cuisine. If there is a tie, return the item with the lexicographically smaller name.

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".

Solution


Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

  1. What are the possible ranges for the food ratings, and how are ratings represented (e.g., integers, floats)?
  2. Can a food item belong to multiple cuisines, and if so, how is that represented in the input?
  3. If a food item with the same name exists in multiple cuisines, how should the system handle that?
  4. When retrieving the highest-rated food in a cuisine, if there's a tie, how should the tie be broken (e.g., alphabetically by food name)?
  5. What is the expected scale of the data (number of foods, number of cuisines), as this may impact data structure choices?

Brute Force Solution

Approach

The brute force method for designing a food rating system involves going through all possible combinations of food, cuisine, and rating updates. We will exhaustively search for the best rated food of a specific cuisine by checking every food item within that cuisine and comparing ratings.

Here's how the algorithm would work step-by-step:

  1. To find the highest-rated food for a specific cuisine, look at every food item in the system.
  2. Check if each food item belongs to the cuisine we are interested in.
  3. If it does, compare its rating to the highest rating we have seen so far for that cuisine.
  4. If the current food item has a higher rating than the previous highest rating, remember the current food item and its rating.
  5. Repeat steps 2-4 for all food items in the system.
  6. After checking every food item, the food item we remembered will be the highest rated food for that cuisine.

Code Implementation

class FoodRatings:

    def __init__(self, foods, cuisines, ratings):
        self.foods = foods
        self.cuisines = cuisines
        self.ratings = ratings

    def changeRating(self, food, newRating):
        for index, current_food in enumerate(self.foods):
            if current_food == food:
                self.ratings[index] = newRating
                break

    def highestRated(self, cuisine):
        highest_rating = -1
        highest_rated_food_for_cuisine = None

        # Iterate through all food items in the system
        for index, current_food in enumerate(self.foods):

            # Check if the current food belongs to the target cuisine
            if self.cuisines[index] == cuisine:

                # Compare current food's rating to the highest rating seen so far
                if self.ratings[index] > highest_rating:
                    highest_rating = self.ratings[index]
                    highest_rated_food_for_cuisine = current_food
                elif self.ratings[index] == highest_rating:
                    #Break Ties by using lexicographical order if the ratings match
                    if current_food < highest_rated_food_for_cuisine:
                        highest_rated_food_for_cuisine = current_food

        return highest_rated_food_for_cuisine

Big(O) Analysis

Time Complexity
O(n)The algorithm iterates through every food item in the system once to find the highest-rated food for a specific cuisine. The input size, n, represents the total number of food items. For each food item, a constant amount of work is performed: checking its cuisine and comparing its rating. Therefore, the time complexity is directly proportional to the number of food items, resulting in O(n).
Space Complexity
O(1)The algorithm described iterates through the input food items to find the highest-rated food for a specific cuisine, but it only stores a constant number of variables: the highest rating seen so far and the corresponding food item. No auxiliary data structures that scale with the input size are created. Therefore, the auxiliary space used is constant, irrespective of the number of food items (N). This means the space complexity is O(1).

Optimal Solution

Approach

To design the food rating system efficiently, we should focus on storing and retrieving information quickly. The key is to use data structures that allow us to easily find the highest-rated food in a specific cuisine without looking at every single food item each time.

Here's how the algorithm would work step-by-step:

  1. First, store all the food, their cuisines, and their ratings in a way that's easy to look up. Think of this like a well-organized phone book for food.
  2. Next, create a separate system that keeps track of the best-rated food for each cuisine. This will let us quickly find the top food for any cuisine.
  3. When a food's rating changes, update both the main food information and the system tracking the best-rated foods.
  4. When someone asks for the highest-rated food in a specific cuisine, use the special system we built for tracking top foods. This avoids looking through all the foods every time.
  5. To make things even faster, consider using a special type of organized list for each cuisine's top foods, which automatically keeps them in order from best to worst.

Code Implementation

class FoodRatingSystem:

    def __init__(self, foods, cuisines, ratings):
        self.food_to_cuisine = {}
        self.food_to_rating = {}
        self.cuisine_to_foods = {}
        for i in range(len(foods)):
            food = foods[i]
            cuisine = cuisines[i]
            rating = ratings[i]
            self.food_to_cuisine[food] = cuisine
            self.food_to_rating[food] = rating

            # Use a sorted list to track top foods efficiently
            if cuisine not in self.cuisine_to_foods:
                self.cuisine_to_foods[cuisine] = []
            self.cuisine_to_foods[cuisine].append((rating, food))
        for cuisine in self.cuisine_to_foods:
            self.cuisine_to_foods[cuisine].sort(key=lambda x: (-x[0], x[1]))

    def changeRating(self, food, new_rating):
        self.food_to_rating[food] = new_rating
        cuisine = self.food_to_cuisine[food]

        # Update rating and sort the list of foods
        foods_in_cuisine = self.cuisine_to_foods[cuisine]
        for i in range(len(foods_in_cuisine)):
            if foods_in_cuisine[i][1] == food:
                foods_in_cuisine[i] = (new_rating, food)
                break
        self.cuisine_to_foods[cuisine].sort(key=lambda x: (-x[0], x[1]))

    def highestRated(self, cuisine):
        # Retrieve highest rated food directly from sorted list
        return self.cuisine_to_foods[cuisine][0][1]

Big(O) Analysis

Time Complexity
O(log n)The food rating system primarily relies on efficient data structures for storing and retrieving information. Assuming a HashMap or Dictionary is used to store food items and their associated data (cuisine and rating), accessing a specific food's information takes O(1) time on average. The most significant operation in terms of complexity is likely managing the highest-rated foods for each cuisine, which we assume is done using a balanced binary search tree (or a similar data structure like a priority queue) for each cuisine. Updating a food's rating would involve removing the old entry and inserting the updated entry into the corresponding cuisine's sorted structure, leading to a time complexity of O(log n) where n is the number of foods in that cuisine.
Space Complexity
O(N)The solution stores food items, their cuisines, and ratings, which requires space proportional to the number of food items. Let N be the total number of food items. A data structure to store the best-rated food for each cuisine is also maintained, potentially requiring space proportional to the number of unique cuisines, which in the worst case could also be N if each food has a unique cuisine. Thus, the auxiliary space used is primarily determined by the need to store information related to each food item, leading to O(N) space complexity.

Edge Cases

CaseHow to Handle
Empty food array or cuisine arrayReturn immediately or throw an exception as no operations can be performed on empty data.
Arrays of food, cuisines, and ratings have different lengthsThrow an IllegalArgumentException since the inputs are inconsistent and will cause errors.
Food names contain special characters or are very longSanitize the food names during input to prevent injection attacks and potential issues with indexing/lookup.
Ratings are negative or zeroThrow an exception or treat them as invalid ratings and skip them, documenting the behaviour.
Cuisines are empty stringsTreat the food as not belonging to any cuisine, or assign it to a default/unspecified cuisine.
Large number of foods and cuisines, potential memory issuesUse efficient data structures (e.g., HashMaps) and consider limiting the number of stored food items or cuisines to prevent memory exhaustion.
Multiple foods with the same name but different cuisine and ratingThe last entered or first entered can be the deciding factor when multiple foods have same name.
Requesting the best rated food for a cuisine that does not existReturn null or a specific error value (e.g., an empty string) indicating the cuisine is not found.