2353. Design a Food Rating 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

/**
 * @param {string[]} foods
 * @param {string[]} cuisines
 * @param {number[]} ratings
 */
var FoodRatings = function(foods, cuisines, ratings) {
    this.foodsMap = {};
    this.cuisinesMap = {};
    this.getRatingWithName = function(food, rating) {
        var z = 'z'.charCodeAt(0);
        var foodNameRating = food.split('')
            .map(char => z - char.charCodeAt(0))
            .map(num => num < 10 ? `0${num}` : `${num}`)
            .join('');
        var rate = Number(`${rating}.${foodNameRating}`);
        return rate;
    };
    for (var i = 0; i < foods.length; i++) {
        var rate = this.getRatingWithName(foods[i], ratings[i]);
        this.foodsMap[foods[i]] = [cuisines[i], rate];
        if (!this.cuisinesMap[cuisines[i]]) {
            this.cuisinesMap[cuisines[i]] = new MaxPriorityQueue();
        }
        this.cuisinesMap[cuisines[i]].enqueue(foods[i], rate);
    }
};

/** 
 * @param {string} food 
 * @param {number} newRating
 * @return {void}
 */
FoodRatings.prototype.changeRating = function(food, newRating) {
    var rate = this.getRatingWithName(food, newRating);
    this.foodsMap[food][1] = rate;
    this.cuisinesMap[this.foodsMap[food][0]].enqueue(food, rate);
};

/**
 * @param {string} cuisine
 * @return {string}
 */
FoodRatings.prototype.highestRated = function(cuisine) {
    var item = this.cuisinesMap[cuisine].front();
    while (this.foodsMap[item.element][1] !== item.priority) {
        this.cuisinesMap[cuisine].dequeue();
        item = this.cuisinesMap[cuisine].front();
    }
    return item.element;
};

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

Explain:

nope.

Complexity:

• Time complexity : O(n * log(n) + m * log(n)).
• Space complexity : O(n).