Design a Food Rating System LeetCode Solution

Problem – Design a Food Rating System LeetCode Solution

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.

Design a Food Rating System LeetCode Solution in C++

class FoodRatings {
public:
    unordered_map<string, set<pair<int, string>>> cuisine_ratings;
    unordered_map<string, string> food_cuisine;
    unordered_map<string, int> food_rating;
    FoodRatings(vector<string>& foods, vector<string>& cuisines, vector<int>& ratings) {
        for (int i = 0; i < foods.size(); ++i) {
            cuisine_ratings[cuisines[i]].insert({ -ratings[i], foods[i] });
            food_cuisine[foods[i]] = cuisines[i];
            food_rating[foods[i]] = ratings[i];
        }
    }
    void changeRating(string food, int newRating) {
        auto &cuisine = food_cuisine.find(food)->second;
        cuisine_ratings[cuisine].erase({ -food_rating[food], food });
        cuisine_ratings[cuisine].insert({ -newRating, food });
        food_rating[food] = newRating;
    }
    string highestRated(string cuisine) {
        return begin(cuisine_ratings[cuisine])->second;
    }
};

Design a Food Rating System LeetCode Solution in Java

HashMap<String, PriorityQueue<Food>> x = new HashMap<>(); // get pq from cuisine name
HashMap<String, Food> menu = new HashMap<>(); // get Food (object) by food name

public FoodRatings(String[] foods, String[] cuisines, int[] ratings) {
    for(int i=0; i<foods.length; i++){
        if(!x.containsKey(cuisines[i])){
            PriorityQueue<Food> pq = new PriorityQueue<>((a,b)->
            b.rating-a.rating==0 ? a.name.compareTo(b.name) : b.rating-a.rating);
            x.put(cuisines[i], pq);
        }
        
        Food curr = new Food(foods[i], cuisines[i], ratings[i]);
		PriorityQueue<Food> pq = x.get(cuisines[i]);
        pq.add(curr);
        menu.put(foods[i], curr);
    }
}

public void changeRating(String food, int newRating) {
    Food curr = menu.get(food);
    PriorityQueue<Food> pq = x.get(curr.cuisine);
    pq.remove(curr);
    curr.rating = newRating;
    pq.add(curr);
}

public String highestRated(String cuisine) {
    return x.get(cuisine).peek().name;
}

class Food{
    int rating;
    String name, cuisine;
    Food(String name, String cuisine, int rating){
        this.name = name; this.rating = rating; this.cuisine = cuisine;
    }
}

Design a Food Rating System LeetCode Solution in Python

class FoodRatings:

    def __init__(self, foods: List[str], cuisines: List[str], ratings: List[int]):
        n = len(foods)
        self.foods = foods; self.cuisines = cuisines; self.ratings = ratings
        self.index = dict(zip(foods, range(n)))
        self.cuisinesMap = defaultdict(list)
        for i, cuisine in enumerate(cuisines):
            heappush(self.cuisinesMap[cuisine], (-self.ratings[i], self.foods[i]))
            
    def changeRating(self, food: str, newRating: int) -> None:
        i = self.index[food]
        self.ratings[i] = newRating; cuisine = self.cuisines[i]
        heappush(self.cuisinesMap[cuisine], (-self.ratings[i], self.foods[i]))

    def highestRated(self, cuisine: str) -> str:
        rating, food = self.cuisinesMap[cuisine][0]
        while rating != -self.ratings[self.index[food]]:  # check highest rating is current
            heappop(self.cuisinesMap[cuisine])
            rating, food = self.cuisinesMap[cuisine][0]
        return food

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