Problem Description

You need to design a food rating system that tracks foods, their cuisines, and ratings. The system should support two main operations: modifying food ratings and finding the highest-rated food for a specific cuisine.

The system is initialized with three arrays of equal length n:

• foods[i] - the name of the i-th food
• cuisines[i] - the cuisine type of the i-th food
• ratings[i] - the initial rating of the i-th food

The FoodRatings class must implement three methods:

1. Constructor FoodRatings(foods, cuisines, ratings): Initializes the system with the given food information.

2. changeRating changeRating(food, newRating): Updates the rating of the specified food to newRating.

3. highestRated highestRated(cuisine): Returns the name of the highest-rated food for the given cuisine. If multiple foods have the same highest rating, return the one that comes first lexicographically (alphabetically).

For example, if you have foods "pizza" and "pasta" both with rating 10 for Italian cuisine, highestRated("Italian") should return "pasta" since it comes before "pizza" alphabetically.

The solution uses two data structures:

• A hash table d that maps each cuisine to a sorted list of (negative rating, food name) tuples. The negative rating is used to maintain descending order by rating.
• A hash table g that maps each food to its (rating, cuisine) pair for quick lookups during rating updates.

When changing a rating, the system removes the old (rating, food) entry from the cuisine's sorted list and adds the new one. Finding the highest-rated food simply returns the first element in the cuisine's sorted list.

Intuition

The key insight is recognizing that we need to efficiently handle two types of queries: updating ratings and finding the highest-rated food per cuisine. This suggests we need fast lookups and ordered data structures.

When thinking about finding the highest-rated food for a cuisine, we need to maintain foods sorted by rating. But since ratings can change, we need a data structure that allows both insertion and deletion while maintaining order - this points us toward using a sorted list or ordered set.

The challenge is that when we update a food's rating, we need to:

1. Know which cuisine it belongs to (to find the right sorted list)
2. Know its current rating (to remove the old entry)
3. Insert the new rating in the correct position

This leads to the idea of maintaining two hash tables:

• One to group foods by cuisine with their ratings in sorted order
• Another to quickly look up a food's current rating and cuisine

For the sorted storage, we use (negative rating, food name) tuples because:

• Python's SortedList sorts in ascending order by default
• We want highest ratings first, so negating the rating achieves descending order
• When ratings are equal, the tuple naturally sorts by the second element (food name) lexicographically

The beauty of this approach is that highestRated becomes a simple O(1) operation - just return the first element in the sorted list for that cuisine. The changeRating operation involves removing one element and adding another to a sorted list, which are both O(log n) operations in a balanced tree structure.

This design trades some space (storing data in two structures) for time efficiency, ensuring all operations are fast even with frequent rating updates.

Learn more about Heap (Priority Queue) patterns.

Solution Approach

The implementation uses two hash tables and a sorted list data structure to efficiently manage the food rating system.

Data Structures:

• self.d: A defaultdict(SortedList) that maps each cuisine to a sorted list of (-rating, food) tuples
• self.g: A dictionary that maps each food to its (rating, cuisine) pair

Constructor Implementation:

The constructor iterates through the input arrays and populates both data structures:

for food, cuisine, rating in zip(foods, cuisines, ratings):
    self.d[cuisine].add((-rating, food))  # Add to cuisine's sorted list
    self.g[food] = (rating, cuisine)       # Store food's info

For each food, we:

1. Add (-rating, food) to the sorted list for its cuisine. The negative rating ensures the list is sorted in descending order by rating.
2. Store the food's rating and cuisine in self.g for quick lookups.

changeRating Method:

When updating a food's rating:

def changeRating(self, food: str, newRating: int) -> None:
    oldRating, cuisine = self.g[food]  # Get current info
    self.g[food] = (newRating, cuisine)  # Update stored rating
    self.d[cuisine].remove((-oldRating, food))  # Remove old entry
    self.d[cuisine].add((-newRating, food))  # Add new entry

The process involves:

1. Retrieve the food's current rating and cuisine from self.g
2. Update the stored rating in self.g to the new value
3. Remove the old (-oldRating, food) tuple from the cuisine's sorted list
4. Insert the new (-newRating, food) tuple into the sorted list

highestRated Method:

Finding the highest-rated food is straightforward:

def highestRated(self, cuisine: str) -> str:
    return self.d[cuisine][0][1]  # Return food name from first tuple

Since the sorted list maintains foods in descending order by rating (and ascending lexicographically for ties), the first element [0] is always the highest-rated food. We return the food name [1] from the tuple.

Time Complexity:

• Constructor: O(n log n) where n is the number of foods
• changeRating: O(log m) where m is the number of foods in the cuisine
• highestRated: O(1)

Space Complexity: O(n) to store all food information in both data structures

Example Walkthrough

Let's walk through a small example to illustrate how the solution works.

Initial Setup:

foods = ["kimchi", "miso", "sushi", "ramen"]
cuisines = ["korean", "japanese", "japanese", "japanese"]
ratings = [9, 12, 8, 15]

Step 1: Constructor Initialization

When we create the FoodRatings object, two data structures are built:

self.d (cuisine → sorted list):

• "korean": [(-9, "kimchi")]
• "japanese": [(-15, "ramen"), (-12, "miso"), (-8, "sushi")]

Note how the Japanese foods are sorted by negative rating, so -15 comes first (highest rating 15).

self.g (food → info):

• "kimchi": (9, "korean")
• "miso": (12, "japanese")
• "sushi": (8, "japanese")
• "ramen": (15, "japanese")

Step 2: Query highestRated("japanese")

Looking at self.d["japanese"], the first element is (-15, "ramen"). Return: "ramen" (the food with rating 15)

Step 3: changeRating("ramen", 10)

1. Look up "ramen" in self.g: get (15, "japanese")
2. Update self.g["ramen"] to (10, "japanese")
3. Remove (-15, "ramen") from self.d["japanese"]
4. Add (-10, "ramen") to self.d["japanese"]

After this operation: self.d["japanese"]: [(-12, "miso"), (-10, "ramen"), (-8, "sushi")]

Step 4: Query highestRated("japanese")

Now the first element is (-12, "miso"). Return: "miso" (rating 12 is now highest)

Step 5: changeRating("sushi", 12)

1. Look up "sushi" in self.g: get (8, "japanese")
2. Update self.g["sushi"] to (12, "japanese")
3. Remove (-8, "sushi") from self.d["japanese"]
4. Add (-12, "sushi") to self.d["japanese"]

After this operation: self.d["japanese"]: [(-12, "miso"), (-12, "sushi"), (-10, "ramen")]

Note that both "miso" and "sushi" have rating 12, but "miso" comes first lexicographically.

Step 6: Query highestRated("japanese")

The first element is still (-12, "miso"). Return: "miso" (when tied at rating 12, "miso" < "sushi" alphabetically)

This example demonstrates how:

• The negative ratings maintain descending order
• Ties are broken lexicographically
• Updates efficiently remove and reinsert elements
• Queries return results in O(1) time

Solution Implementation

Time and Space Complexity

Time Complexity:

• Constructor (__init__): O(n log n) where n is the number of foods. The constructor iterates through all foods once with O(n) iteration, and for each food, it performs an insertion into a SortedList which takes O(log k) time where k is the current size of the SortedList for that cuisine. In the worst case where all foods belong to the same cuisine, this becomes O(n log n).

• changeRating method: O(log n) in the worst case. This method performs:

  • Dictionary lookup in self.g: O(1)
  • Dictionary update in self.g: O(1)
  • Removal from SortedList: O(log k) where k is the number of foods in that cuisine
  • Addition to SortedList: O(log k) where k is the number of foods in that cuisine
  • In the worst case where all foods belong to the same cuisine, k = n, so the complexity is O(log n)

• highestRated method: O(1). This method simply accesses the first element of the SortedList for the given cuisine, which is a constant-time operation.

Space Complexity: O(n) where n is the number of foods.

• self.d stores all foods organized by cuisine in SortedLists: O(n) total space
• self.g stores a tuple (rating, cuisine) for each food: O(n) space
• Total space complexity: O(n)

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Forgetting to Handle Lexicographical Ordering for Ties

One of the most common mistakes is not properly handling the case where multiple foods have the same rating. The problem requires returning the lexicographically smallest food name when there's a tie.

Incorrect Approach:

# Wrong: Only storing rating without considering food name for ties
self.cuisine_to_foods[cuisine].add((-rating,))  # Missing food name

Correct Solution:

# Store both rating and food name as a tuple
self.cuisine_to_foods[cuisine].add((-rating, food))

The SortedList will automatically sort tuples first by the negative rating, then by food name alphabetically when ratings are equal.

2. Using Positive Ratings Instead of Negative

Another frequent error is storing positive ratings in the sorted list, which would sort in ascending order (lowest ratings first) rather than the required descending order.

Incorrect Approach:

# Wrong: This sorts in ascending order by rating
self.cuisine_to_foods[cuisine].add((rating, food))
# highestRated would incorrectly return the lowest-rated food

Correct Solution:

# Use negative rating to achieve descending order
self.cuisine_to_foods[cuisine].add((-rating, food))

3. Not Updating Both Data Structures in changeRating

A critical mistake is forgetting to update the food_info dictionary when changing a rating, which would cause future rating changes to use outdated information.

Incorrect Approach:

def changeRating(self, food: str, newRating: int) -> None:
    old_rating, cuisine = self.food_info[food]
    # Forgot to update food_info!
    self.cuisine_to_foods[cuisine].remove((-old_rating, food))
    self.cuisine_to_foods[cuisine].add((-newRating, food))

Correct Solution:

def changeRating(self, food: str, newRating: int) -> None:
    old_rating, cuisine = self.food_info[food]
    # Must update food_info with new rating
    self.food_info[food] = (newRating, cuisine)
    self.cuisine_to_foods[cuisine].remove((-old_rating, food))
    self.cuisine_to_foods[cuisine].add((-newRating, food))

4. Attempting to Remove Non-existent Entries

When implementing changeRating, ensure you're removing the exact tuple that exists in the sorted list. Any mismatch in the tuple structure will cause a ValueError.

Incorrect Approach:

# Wrong: Trying to remove with positive rating when stored as negative
self.cuisine_to_foods[cuisine].remove((old_rating, food))  # ValueError!

Correct Solution:

# Remove with the same format used when adding
self.cuisine_to_foods[cuisine].remove((-old_rating, food))