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Formatted question description: https://leetcode.ca/all/2353.html

2353. Design a Food Rating System

  • Difficulty: Medium.
  • Related Topics: Hash Table, Design, Heap (Priority Queue), Ordered Set.
  • Similar Questions: Design a Number Container System.

Problem

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 (Java, C++, Python)

  • class FoodRatings {
    private HashMap<String, TreeSet<Food>> cus = new HashMap<>();
    private HashMap<String, Food> foodHashMap = new HashMap<>();

    public FoodRatings(String[] foods, String[] cuisines, int[] ratings) {
        for (int i = 0; i < foods.length; i++) {
            Food food = new Food(foods[i], ratings[i], cuisines[i]);
            foodHashMap.put(foods[i], food);
            if (cus.containsKey(cuisines[i])) {
                cus.get(cuisines[i]).add(food);
            } else {
                TreeSet<Food> pq = new TreeSet<>(new Comp());
                pq.add(food);
                cus.put(cuisines[i], pq);
            }
        }
    }

    public void changeRating(String food, int newRating) {
        Food dish = foodHashMap.get(food);
        TreeSet<Food> pq = cus.get(dish.cus);
        pq.remove(dish);
        dish.rating = newRating;
        pq.add(dish);
    }

    public String highestRated(String cuisine) {
        return cus.get(cuisine).first().foodItem;
    }

    private static class Comp implements Comparator<Food> {
        public int compare(Food f1, Food f2) {
            if (f1.rating == f2.rating) {
                return f1.foodItem.compareTo(f2.foodItem);
            }
            return Integer.compare(f2.rating, f1.rating);
        }
    }

    private static class Food {
        private String foodItem;
        private String cus;
        private int rating;

        Food(String food, int rating, String cus) {
            this.foodItem = food;
            this.rating = rating;
            this.cus = cus;
        }
    }
}

/**
 * Your FoodRatings object will be instantiated and called as such:
 * FoodRatings obj = new FoodRatings(foods, cuisines, ratings);
 * obj.changeRating(food,newRating);
 * String param_2 = obj.highestRated(cuisine);
 */

  • from sortedcontainers import SortedSet


class FoodRatings:
    def __init__(self, foods: List[str], cuisines: List[str], ratings: List[int]):
        self.mp = {}
        self.t = defaultdict(lambda: SortedSet(key=lambda x: (-x[0], x[1])))

        for a, b, c in zip(foods, cuisines, ratings):
            self.mp[a] = (b, c)
            self.t[b].add((c, a))

    def changeRating(self, food: str, newRating: int) -> None:
        b, c = self.mp[food]
        self.mp[food] = (b, newRating)
        self.t[b].remove((c, food))
        self.t[b].add((newRating, food))

    def highestRated(self, cuisine: str) -> str:
        return self.t[cuisine][0][1]


# Your FoodRatings object will be instantiated and called as such:
# obj = FoodRatings(foods, cuisines, ratings)
# obj.changeRating(food,newRating)
# param_2 = obj.highestRated(cuisine)

############

# 2353. Design a Food Rating System
# https://leetcode.com/problems/design-a-food-rating-system/

from sortedcontainers import SortedList

class FoodRatings:

    def __init__(self, foods: List[str], cuisines: List[str], ratings: List[int]):
        self.mp = {}
        self.A = defaultdict(SortedList)
        
        for food, cuisine, rating in zip(foods, cuisines, ratings):
            self.mp[food] = (cuisine, rating)
            self.A[cuisine].add((-rating, food))
        
    def changeRating(self, food: str, newRating: int) -> None:
        cuisine, rating = self.mp[food]
        self.mp[food] = (cuisine, newRating)
        self.A[cuisine].remove((-rating, food))
        self.A[cuisine].add((-newRating, food))

    def highestRated(self, cuisine: str) -> str:
        return self.A[cuisine][0][1]


# Your FoodRatings object will be instantiated and called as such:
# obj = FoodRatings(foods, cuisines, ratings)
# obj.changeRating(food,newRating)
# param_2 = obj.highestRated(cuisine)


  • using pis = pair<int, string>;

class FoodRatings {
    map<string, pis> mp;
    map<string, set<pis>> t;

public:
    FoodRatings(vector<string>& foods, vector<string>& cuisines, vector<int>& ratings) {
        int n = foods.size();
        for (int i = 0; i < n; ++i) {
            string a = foods[i], b = cuisines[i];
            int c = ratings[i];
            mp[a] = pis(c, b);
            t[b].insert(pis(-c, a));
        }
    }

    void changeRating(string food, int newRating) {
        pis& p = mp[food];
        t[p.second].erase(pis(-p.first, food));
        p.first = newRating;
        t[p.second].insert(pis(-p.first, food));
    }

    string highestRated(string cuisine) {
        return t[cuisine].begin()->second;
    }
};

/**
 * Your FoodRatings object will be instantiated and called as such:
 * FoodRatings* obj = new FoodRatings(foods, cuisines, ratings);
 * obj->changeRating(food,newRating);
 * string param_2 = obj->highestRated(cuisine);
 */

Explain:

nope.

Complexity:

  • Time complexity : O(n).
  • Space complexity : O(n).

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