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Design a Food Rating System

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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.

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 ith food,
    • cuisines[i] is the type of cuisine of the ith food, and
    • ratings[i] is the initial rating of the ith 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.

Note that a string x is lexicographically smaller than string y if x comes before y in dictionary order, that is, either x is a prefix of y, or if i is the first position such that x[i] != y[i], then x[i] comes before y[i] in alphabetic order.

Example 1:

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


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. Regarding the `changeRating` method, are there any constraints on the `newRating` value? For instance, can it be zero, negative, or does it have to conform to the same positive range as the initial ratings?
  2. Once a food item is initialized with a cuisine, is that relationship fixed? In other words, will a food item like 'sushi' always be associated with the 'japanese' cuisine throughout all operations?
  3. The problem states that all food names in the initial input are distinct. Does this uniqueness guarantee hold for the lifetime of the system? This is important for choosing a primary key for our data structures.
  4. What is the expected behavior if the `highestRated` method is called for a cuisine that was provided initially but for which we have no current food items (hypothetically, if a removal operation existed)? The prompt guarantees the cuisine exists, but is it also guaranteed to be non-empty at the time of the call?
  5. Considering the potential for a large number of unique cuisines and food items, are there any memory constraints I should be aware of, or is the primary goal to optimize for the time complexity of the `changeRating` and `highestRated` methods?

Brute Force Solution

Approach

The straightforward approach is to maintain a single, master list of all foods with their details. To find the best food for a certain cuisine, we simply check every single food in the entire list, one by one, to see which one is the best fit according to the rules.

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

  1. First, create one big collection to store the name, cuisine type, and rating for every food item.
  2. When a food's rating needs to be changed, search through this entire collection from start to finish until you find that specific food.
  3. Once you find it, update its rating to the new value.
  4. When asked to find the highest-rated food for a specific cuisine, you'll start a fresh search through the entire collection.
  5. For each food, first check if it's the type of cuisine you're looking for.
  6. If it is, compare it to the best food you've found so far for that cuisine.
  7. A food is considered 'better' if its rating is higher. If the ratings are the same, the food whose name comes first alphabetically is better.
  8. Continue this process for every single food in your collection.
  9. After you've checked them all, the food you determined to be the 'best so far' is the one you report as the answer.

Code Implementation

class FoodRatings:
    def __init__(self, foods: list[str], cuisines: list[str], ratings: list[int]):
        # Create a single master list to hold all food details for straightforward iteration.
        self.food_master_list = []
        for index in range(len(foods)):
            food_name = foods[index]
            cuisine_type = cuisines[index]
            initial_rating = ratings[index]
            self.food_master_list.append([food_name, cuisine_type, initial_rating])

    def changeRating(self, food: str, newRating: int) -> None:
        # We must search the entire list to find the matching food by its name to update the rating.
        for food_record in self.food_master_list:
            if food_record[0] == food:
                food_record[2] = newRating
                return

    def highestRated(self, cuisine: str) -> str:
        # Initialize a variable to keep track of the best food found so far for the target cuisine.
        current_best_food_for_cuisine = None

        for food_record in self.food_master_list:
            food_name = food_record[0]
            food_cuisine = food_record[1]
            food_rating = food_record[2]

            if food_cuisine == cuisine:
                # A food becomes the new winner if its rating is higher, or if tied, its name is smaller.
                if current_best_food_for_cuisine is None or \
                   food_rating > current_best_food_for_cuisine[2] or \
                   (food_rating == current_best_food_for_cuisine[2] and food_name < current_best_food_for_cuisine[0]):
                    current_best_food_for_cuisine = food_record
        
        return current_best_food_for_cuisine[0]

Big(O) Analysis

Time Complexity
O(n)Let n be the total number of food items in the master list. The critical operations, `changeRating` and `highestRated`, both rely on a linear scan through this entire list. To change a rating or find the highest rated food, we must potentially examine every single one of the n food items. This leads to a total number of operations directly proportional to n for each function call, which simplifies to a time complexity of O(n).
Space Complexity
O(N)The solution's space complexity is determined by the 'one big collection' used to store all food data. If we define N as the total number of food items, this collection must store N distinct entries, each containing a name, cuisine, and rating. While the search operations use a few temporary variables to keep track of the best food found so far, this is constant extra space. Therefore, the dominant factor for memory usage is the main collection, which grows linearly with the number of foods, N.

Optimal Solution

Approach

The optimal strategy avoids searching through all the foods every time. Instead, it maintains a separate, continuously sorted 'leaderboard' for each cuisine, making it instantly possible to find the top-rated food for any given cuisine.

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

  1. First, create a master directory where you can look up any food by its name to quickly find its current rating and what type of cuisine it is.
  2. Next, for each individual cuisine, create its own dedicated leaderboard. Think of it as a ranked list of all foods belonging to that cuisine.
  3. Each leaderboard must always be kept perfectly sorted. The primary sorting rule is by the highest rating, and if there's a tie in rating, the food with the name that comes first alphabetically wins the higher spot.
  4. When asked for the highest-rated food of a certain cuisine, the answer is simple: just look at the specific leaderboard for that cuisine and pick the food at the very top.
  5. When a food's rating changes, the system needs to update its rank. First, use the master directory to find the food's old rating and its cuisine.
  6. Then, go to the correct cuisine's leaderboard, remove the food's old entry, and add it back with its new, updated rating. The leaderboard will automatically place it in its new correct sorted position.
  7. Finally, remember to also update the food's new rating in the master directory to keep everything in sync for future changes.

Code Implementation

class FoodRatings:
    def __init__(self, foods: list[str], cuisines: list[str], ratings: list[int]):
        # A map for O(1) access to a food's current rating and cuisine.
        self.food_info = {}

        self.cuisine_leaderboards = collections.defaultdict(list)

        for index in range(len(foods)):
            food_name = foods[index]
            cuisine_name = cuisines[index]
            initial_rating = ratings[index]
            
            self.food_info[food_name] = [initial_rating, cuisine_name]
            # Storing ratings as negative allows a min-heap to function as a max-heap.
            heapq.heappush(self.cuisine_leaderboards[cuisine_name], (-initial_rating, food_name))

    def changeRating(self, food_to_update: str, new_rating: int) -> None:
        original_cuisine = self.food_info[food_to_update][1]
        self.food_info[food_to_update][0] = new_rating
        
        # This lazy-add avoids a costly search-and-remove from the heap for the old rating.
        heapq.heappush(self.cuisine_leaderboards[original_cuisine], (-new_rating, food_to_update))

    def highestRated(self, cuisine_to_query: str) -> str:
        leaderboard_for_cuisine = self.cuisine_leaderboards[cuisine_to_query]
        
        # Stale ratings might be at the top, so we must find the first valid one.
        while -leaderboard_for_cuisine[0][0] != self.food_info[leaderboard_for_cuisine[0][1]][0]:
            heapq.heappop(leaderboard_for_cuisine)
        
        return leaderboard_for_cuisine[0][1]

Big(O) Analysis

Time Complexity
O(log n)The time complexity is determined by the changeRating operation, as it is the most complex repeated action. The cost is driven by updating the continuously sorted leaderboard for a given cuisine. This update requires removing the food's old rating and inserting the new one into a data structure that maintains sorted order, like a balanced binary search tree. Both removal and insertion in such a structure take logarithmic time relative to its size, leading to a complexity of O(log n), where n is the total number of foods in the worst-case.
Space Complexity
O(N)The solution uses two main data structures where N is the total number of initial foods. A 'master directory', likely a hash map, is created to store the cuisine and rating for each of the N foods. A second hash map is used to hold a dedicated sorted data structure, or 'leaderboard', for each cuisine. Cumulatively, these leaderboards also store information for all N foods. Since the storage for both of these structures grows linearly with the number of foods, the total auxiliary space complexity is O(N).

Edge Cases

Multiple food items in a cuisine share the same maximum rating.
How to Handle:
A custom comparator for the priority queue ensures ties in rating are broken by choosing the lexicographically smaller food name.
A single food item's rating is changed multiple times consecutively.
How to Handle:
The lazy deletion strategy causes outdated entries to build up in the heap, which are filtered out during the next relevant `highestRated` query.
A single cuisine contains almost all food items, while other cuisines contain only one or a few.
How to Handle:
Performance for `highestRated` is proportional to the logarithm of the number of foods in the queried cuisine, gracefully handling the skewed distribution.
The system is initialized with the maximum number of foods and receives the maximum number of method calls.
How to Handle:
The use of hash maps and heaps provides logarithmic time complexity for operations, ensuring the solution scales to meet the maximum constraints.
The system is initialized with only one food item for a given cuisine.
How to Handle:
The data structures handle the base case of a single item correctly, with `highestRated` for its cuisine always returning that one item.
changeRating is called with a newRating that is identical to the food's current rating.
How to Handle:
The logic still functions correctly by adding a new entry to the heap, which does not alter the correctness of `highestRated` calls.
Food names involved in a tie-breaking situation have different lengths or are prefixes of each other.
How to Handle:
The solution leverages built-in string comparison functions which properly handle prefixes and different lengths according to standard lexicographical rules.
A rating is changed to a lower value, potentially making a different food item the new highest rated.
How to Handle:
The new, lower-rated entry is added to the heap, and the old, higher-rated entry is invalidated and will be ignored by future `highestRated` calls.