LeetCode #2353 — MEDIUM

Design a Food Rating System

Move from brute-force thinking to an efficient approach using array strategy.

Solve on LeetCode
The Problem

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.
Patterns Used

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: 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

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map · Design · Segment Tree

Example 1

["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"]]

Related Problems

  • Design a Number Container System (design-a-number-container-system)
  • Most Popular Video Creator (most-popular-video-creator)
Step 02

Core Insight

What unlocks the optimal approach

  • The key to solving this problem is to properly store the data using the right data structures.
  • Firstly, a hash table is needed to efficiently map each food item to its cuisine and current rating.
  • In addition, another hash table is needed to map cuisines to foods within each cuisine stored in an ordered set according to their ratings.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. Continue until all input is consumed.
Use the first example testcase as your mental trace to verify each transition.
Step 04

Edge Cases

Minimum Input
Single element / shortest valid input
Validate boundary behavior before entering the main loop or recursion.
Duplicates & Repeats
Repeated values / repeated states
Decide whether duplicates should be merged, skipped, or counted explicitly.
Extreme Constraints
Upper-end input sizes
Re-check complexity target against constraints to avoid time-limit issues.
Invalid / Corner Shape
Empty collections, zeros, or disconnected structures
Handle special-case structure before the core algorithm path.
Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2353: Design a Food Rating System
class FoodRatings {
    private Map<String, TreeSet<Pair<Integer, String>>> d = new HashMap<>();
    private Map<String, Pair<Integer, String>> g = new HashMap<>();
    private final Comparator<Pair<Integer, String>> cmp = (a, b) -> {
        if (!a.getKey().equals(b.getKey())) {
            return b.getKey().compareTo(a.getKey());
        }
        return a.getValue().compareTo(b.getValue());
    };

    public FoodRatings(String[] foods, String[] cuisines, int[] ratings) {
        for (int i = 0; i < foods.length; ++i) {
            String food = foods[i], cuisine = cuisines[i];
            int rating = ratings[i];
            d.computeIfAbsent(cuisine, k -> new TreeSet<>(cmp)).add(new Pair<>(rating, food));
            g.put(food, new Pair<>(rating, cuisine));
        }
    }

    public void changeRating(String food, int newRating) {
        Pair<Integer, String> old = g.get(food);
        int oldRating = old.getKey();
        String cuisine = old.getValue();
        g.put(food, new Pair<>(newRating, cuisine));
        d.get(cuisine).remove(new Pair<>(oldRating, food));
        d.get(cuisine).add(new Pair<>(newRating, food));
    }

    public String highestRated(String cuisine) {
        return d.get(cuisine).first().getValue();
    }
}

/**
 * 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);
 */
Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(1) per op
Space
O(n)

Approach Breakdown

NAIVE
O(n) per op time
O(n) space

Use a simple list or array for storage. Each operation (get, put, remove) requires a linear scan to find the target element — O(n) per operation. Space is O(n) to store the data. The linear search makes this impractical for frequent operations.

OPTIMIZED DESIGN
O(1) per op time
O(n) space

Design problems target O(1) amortized per operation by combining data structures (hash map + doubly-linked list for LRU, stack + min-tracking for MinStack). Space is always at least O(n) to store the data. The challenge is achieving constant-time operations through clever structure composition.

Shortcut: Combine two data structures to get O(1) for each operation type. Space is always O(n).
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.