What data structures are used to solve Design a Food Rating System?

This problem uses Array, Hash Table, String, Design, Heap (Priority Queue), Ordered Set. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Design a Food Rating System?

Insight 1: Hash tables are crucial for efficient lookups of food ratings and cuisines. Insight 2: Priority queues (heaps) are ideal for maintaining the highest-rated food for each cuisine.

Which companies ask LeetCode 2353. Design a Food Rating System in interviews?

This problem is frequently asked at Altimetrik, Atlassian.

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Design a Food Rating System - LeetCode 2353 Solution

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

The optimal solution has a time complexity of O(log N) for `changeRating` and O(log N) for `highestRated` in the worst case where N is the number of foods for a cuisine. `changeRating` pushes an element to the heap. `highestRated` might poll several outdated elements before finding the valid highest rated food, but each poll operation is O(log N). Initialization is O(N log N) if we insert into heap. If we just store the elements into the heap and then heapify it, it'll be O(N)..

Q: What is the space complexity of Design a Food Rating System?

The space complexity is O(N), where N is the number of foods. The space is used to store the `food_to_rating`, `food_to_cuisine`, and `cuisine_heaps`. In the worst case, all foods belong to the same cuisine, leading to a priority queue of size N..

Design a Food Rating System - Complete Solution Guide

Design a Food Rating System is LeetCode problem 2353, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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. 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 th food, cuisines[i] is the type of cuisine of the i th food, and rating

Detailed Explanation

The problem requires designing a food rating system. This system should allow you to: 1) Modify the rating of a given food and 2) Retrieve the highest-rated food for a specific cuisine. When multiple foods of the same cuisine have the same highest rating, the food item that is lexicographically smallest should be returned. The input consists of three arrays: `foods` (names of foods), `cuisines` (corresponding cuisine for each food), and `ratings` (initial ratings for each food). The system should be initialized with these arrays and should respond to `changeRating` and `highestRated` queries.

Solution Approach

The solution utilizes hash tables and priority queues. A hash table (`food_to_rating`) stores the current rating of each food item for fast updates. Another hash table (`food_to_cuisine`) stores the cuisine of each food. A hash table of priority queues (`cuisine_heaps`) stores a max-heap for each cuisine, containing tuples of `(-rating, food)`, using negative rating to simulate max-heap behavior with min-heaps available in Python's `heapq` module. When the `changeRating` function is called, the corresponding rating is updated in `food_to_rating`, and a new entry with the updated rating is added to the cuisine's heap. `highestRated` uses lazy deletion. Instead of deleting entries from the priority queue when ratings change, it allows obsolete entries to remain. When peeking at the top element, it checks if that food's rating matches the stored rating in `food_to_rating`. If there is a mismatch, the element is popped, indicating it is stale. The algorithm continues until it finds a valid, up-to-date top element.

Step-by-Step Algorithm

Step 1: Initialization: Create hash tables (`food_to_rating`, `food_to_cuisine`, and `cuisine_heaps`).
Step 2: Populate the hash tables and priority queues with initial data. For each food, store its rating and cuisine, and push `(-rating, food)` onto the appropriate cuisine's heap.
Step 3: `changeRating(food, newRating)`: Update the `food_to_rating` map with the `newRating` and push `(-newRating, food)` into the cuisine's heap.
Step 4: `highestRated(cuisine)`: Iterate through the heap for given cuisine. For each entry, check if its stored rating matches the current rating in `food_to_rating`. If they match, return food. If not, pop the top and repeat until a valid food or the heap is empty.

Key Insights

Insight 1: Hash tables are crucial for efficient lookups of food ratings and cuisines.
Insight 2: Priority queues (heaps) are ideal for maintaining the highest-rated food for each cuisine.
Insight 3: Lazy deletion using ratings map can avoid unnecessary heap updates and maintain heap efficiency.

Complexity Analysis

Time Complexity: O(log N) for `changeRating` and O(log N) for `highestRated` in the worst case where N is the number of foods for a cuisine. `changeRating` pushes an element to the heap. `highestRated` might poll several outdated elements before finding the valid highest rated food, but each poll operation is O(log N). Initialization is O(N log N) if we insert into heap. If we just store the elements into the heap and then heapify it, it'll be O(N).

Space Complexity: O(N), where N is the number of foods. The space is used to store the `food_to_rating`, `food_to_cuisine`, and `cuisine_heaps`. In the worst case, all foods belong to the same cuisine, leading to a priority queue of size N.

Topics

This problem involves: Array, Hash Table, String, Design, Heap (Priority Queue), Ordered Set.

Companies

Asked at: Altimetrik, Atlassian.

Design a Food Rating System | LeetCode 2353

Medium

Array Hash Table String Design Heap (Priority Queue)Ordered Set

Time: O(log N) for `changeRating` and O(log N) for `highestRated` in the worst case where N is the number of foods for a cuisine. `changeRating` pushes an element to the heap. `highestRated` might poll several outdated elements before finding the valid highest rated food, but each poll operation is O(log N). Initialization is O(N log N) if we insert into heap. If we just store the elements into the heap and then heapify it, it'll be O(N).Space: O(N), where N is the number of foods. The space is used to store the `food_to_rating`, `food_to_cuisine`, and `cuisine_heaps`. In the worst case, all foods belong to the same cuisine, leading to a priority queue of size N.

Solution Code

Available in:

import java.util.*;

class FoodRatings {

    private Map<String, int[]> foodInfo;
    private Map<String, PriorityQueue<FoodRating>> cuisineFoods;

    public FoodRatings(String[] foods, String[] cuisines, int[] ratings) {
        foodInfo = new HashMap<>();
        cuisineFoods = new HashMap<>();

        for (int i = 0; i < foods.length; i++) {
            String food = foods[i];
            String cuisine = cuisines[i];
            int rating = ratings[i];

            foodInfo.put(food, new int[]{rating, i}); // Store rating and original index
            cuisineFoods.computeIfAbsent(cuisine, k -> new PriorityQueue<>(
                    (a, b) -> {
                        int ratingComparison = Integer.compare(b.rating, a.rating);
                        if (ratingComparison != 0) {
                            return ratingComparison;
                        } else {
                            return a.food.compareTo(b.food);
                        }
                    }
            )).add(new FoodRating(food, rating));
        }
    }

    public void changeRating(String food, int newRating) {
        int[] info = foodInfo.get(food);
        String cuisine = "";
        
        for (Map.Entry<String, PriorityQueue<FoodRating>> entry : cuisineFoods.entrySet()) {
            PriorityQueue<FoodRating> pq = entry.getValue();
            for (FoodRating fr : pq) {
                if(fr.food.equals(food)){
                    cuisine = entry.getKey();
                    break;
                }
            }
             if (!cuisine.isEmpty()) {
                break;
            }
        }
        

        foodInfo.put(food, new int[]{newRating, foodInfo.get(food)[1]});

        cuisineFoods.get(cuisine).add(new FoodRating(food, newRating));
    }

    public String highestRated(String cuisine) {
        PriorityQueue<FoodRating> heap = cuisineFoods.get(cuisine);

        while (!heap.isEmpty()) {
            FoodRating foodRating = heap.peek();
            String food = foodRating.food;
            int rating = foodRating.rating;
            
            if (foodInfo.containsKey(food) && foodInfo.get(food)[0] == rating) {
                return food;
            }
            heap.poll();
        }
        return null; 
    }

    private static class FoodRating {
        String food;
        int rating;

        public FoodRating(String food, int rating) {
            this.food = food;
            this.rating = rating;
        }
    }
}

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Problem Info

Description

Problem 2353: Design a Food Rating System - Detailed explanation and solution approaches.

Topics

Array Hash Table String Design Heap (Priority Queue)Ordered Set

Companies

AltimetrikAtlassian

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:2353

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

4 steps

1
Step 1: Initialization: Create hash tables (`food_to_rating`, `food_to_cuisine`, and `cuisine_heaps`).
2
Step 2: Populate the hash tables and priority queues with initial data. For each food, store its rating and cuisine, and push `(-rating, food)` onto the appropriate cuisine's heap.
3
Step 3: `changeRating(food, newRating)`: Update the `food_to_rating` map with the `newRating` and push `(-newRating, food)` into the cuisine's heap.
4
Step 4: `highestRated(cuisine)`: Iterate through the heap for given cuisine. For each entry, check if its stored rating matches the current rating in `food_to_rating`. If they match, return food. If not, pop the top and repeat until a valid food or the heap is empty.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(log N) for `changeRating` and O(log N) for `highestRated` in the worst case where N is the number of foods for a cuisine. `changeRating` pushes an element to the heap. `highestRated` might poll several outdated elements before finding the valid highest rated food, but each poll operation is O(log N). Initialization is O(N log N) if we insert into heap. If we just store the elements into the heap and then heapify it, it'll be O(N).

🚀 Logarithmic - Very efficient! Halves the problem each step.

💾

Space Complexity

O(N), where N is the number of foods. The space is used to store the `food_to_rating`, `food_to_cuisine`, and `cuisine_heaps`. In the worst case, all foods belong to the same cuisine, leading to a priority queue of size N.

📝

Why?

The `changeRating` operation involves updating a hash map (O(1)) and pushing an element onto a heap (O(log N)). The `highestRated` operation, in the worst case, might have to remove several obsolete entries from the heap before finding a valid entry, taking O(K log N) where K is the number of stale elements which is bounded by total operations. However, amortized time complexity is still O(log N), because it is guaranteed that at least one valid result exists for each call. The space is linear because we store each food's rating, cuisine, and references to food in the heaps.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Ordered Set

Use an ordered set (e.g., TreeMap in Java) for each cuisine. The ordered set stores the foods with their ratings, sorted by rating (descending) and then lexicographically by food name. `changeRating` would require removing and re-inserting into the ordered set, which may be more cumbersome than the heap-based lazy deletion approach.

Time: O(log N) for `changeRating` and O(log N) for `highestRated`, where N is the number of foods in the cuisine.Space: O(N)

Trade-offs: Ordered set approach might be slightly more complex to implement due to the need to remove and re-insert on rating changes. The heap-based approach with lazy deletion is generally cleaner and more efficient in this scenario.

✨

Key Insights

1Insight 1: Hash tables are crucial for efficient lookups of food ratings and cuisines.
2Insight 2: Priority queues (heaps) are ideal for maintaining the highest-rated food for each cuisine.
3Insight 3: Lazy deletion using ratings map can avoid unnecessary heap updates and maintain heap efficiency.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: Multiple foods in the same cuisine have the same highest rating. The solution uses string comparison to return the lexicographically smaller food.
Edge case 2: A food's rating is changed multiple times before `highestRated` is called for its cuisine. The lazy deletion mechanism in `highestRated` ensures that only the most up-to-date rating is considered.
Edge case 3: `highestRated` is called on a cuisine that has no foods. The solution handles this by iterating until heap is empty and returns an empty string. The constraints ensure that this won't happen.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the lexicographical tie-breaking condition correctly when multiple foods have the same highest rating.

✗Common mistake 2: Naively updating the heaps every time a rating changes, which can be inefficient. The lazy deletion mechanism avoids this issue.

✗Common mistake 3: Forgetting to handle the case where `highestRated` might need to remove multiple stale entries before finding the actual highest-rated food.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Restaurant recommendation systems. The system could track user ratings of dishes from different cuisines and use this algorithm to recommend the highest-rated dish for a specific cuisine.

🏦Real-world scenario 2: E-commerce product rating system. Products can be categorized (similar to cuisines) and the algorithm can be used to quickly find the highest-rated product within a given category.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: June 14, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 13

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 6

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Design a Food Rating System"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether hash table techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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