Leetcode solution 2353. Design a Food Rating System

 

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 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.

Problem link

 

Video Tutorial

You can find the detailed video tutorial here

• Youtube
• B站 

 

Thought Process

If we want to optimize highestRated API call (which we should), a max heap would help achieve it with O(1). Some details about implementation. 

• Make each food into a class would make code clean, also maintain the order when insert/remove from the max heap through Comparable interface. 
• Make sure we have O(1) access to each Food object by maintaining two HashMaps. 
• foodIndex, Map<String, Food> -> Key: food name, Value: Food object 
• CuisineIndex, Map<String, Queue<Food>>, Key: cuisine name, Value: Max heap on rating 
• Follow up thought: if this is a system design question, how would you actually do it? Some read on how Youtube calculates total views.

 

Solutions

 

import java.util.HashMap;
import java.util.Map;
import java.util.PriorityQueue;
import java.util.Queue;

public class FoodRatings {
    
    private class Food implements Comparable<Food>{
        // skip boilerplate getter/setters
        public String food;
        public String cuisine;
        public int rating;

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

        @Override
        public int compareTo(Food o) {
            assert o != null;
            if (o.rating == this.rating) {
                return this.food.compareTo(o.food);
            }
            return o.rating - this.rating; }
    }

    // food -> Food
    private Map<String, Food> foodIndex = new HashMap<>();
    // cuisine -> MaxHeap on rating
    private Map<String, Queue<Food>> cuisineIndex = new HashMap<>();

    public FoodRatings(String[] foods, String[] cuisines, int[] ratings) {
        for (int i = 0; i < foods.length; i++) {
            Food f = new Food(foods[i], cuisines[i], ratings[i]);
            // given food is distinct

            this.foodIndex.put(f.food, f);
            /* use putIfAbsent instead
            if (!this.cuisineIndex.containsKey(f.cuisine)) {
                this.cuisineIndex.put(f.cuisine, new PriorityQueue<>());
            }
            */
            this.cuisineIndex.putIfAbsent(f.cuisine, new PriorityQueue<>());
            this.cuisineIndex.get(f.cuisine).add(f);
        }
    }

    public void changeRating(String food, int newRating) {
        Food f = this.foodIndex.get(food);
        f.rating = newRating;
        // need to update the max heap by deleting and re-inserting
        this.cuisineIndex.get(f.cuisine).remove(f);
        this.cuisineIndex.get(f.cuisine).add(f);
    }

    public String highestRated(String cuisine) {
        return this.cuisineIndex.get(cuisine).peek().food;
    }

}

 

 

Time Complexity: highestRated: O(1), changeRating: O(N) because we have to remove the object, and in order to find the object in a max heap, it is O(N) scan (even though re-insert is O(lgN)). 

Space Complexity: O(N) because we used extra maps linear to food names (which is larger than cuisine names)

 

References

• Task Scheduler using heap
• LRU Cache using heap