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.
Move from brute-force thinking to an efficient approach using array strategy.
Design a food rating system that can do the following:
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, andratings[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 * 104n == foods.length == cuisines.length == ratings.length1 <= foods[i].length, cuisines[i].length <= 10foods[i], cuisines[i] consist of lowercase English letters.1 <= ratings[i] <= 108foods 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.2 * 104 calls in total will be made to changeRating and highestRated.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
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array · Hash Map · Design · Segment Tree
["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"]]
design-a-number-container-system)most-popular-video-creator)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);
*/
// Accepted solution for LeetCode #2353: Design a Food Rating System
import (
"github.com/emirpasic/gods/v2/trees/redblacktree"
)
type pair struct {
rating int
food string
}
type FoodRatings struct {
d map[string]*redblacktree.Tree[pair, struct{}]
g map[string]pair
}
func Constructor(foods []string, cuisines []string, ratings []int) FoodRatings {
d := make(map[string]*redblacktree.Tree[pair, struct{}])
g := make(map[string]pair)
for i, food := range foods {
rating, cuisine := ratings[i], cuisines[i]
g[food] = pair{rating, cuisine}
if d[cuisine] == nil {
d[cuisine] = redblacktree.NewWith[pair, struct{}](func(a, b pair) int {
return cmp.Or(b.rating-a.rating, strings.Compare(a.food, b.food))
})
}
d[cuisine].Put(pair{rating, food}, struct{}{})
}
return FoodRatings{d, g}
}
func (this *FoodRatings) ChangeRating(food string, newRating int) {
p := this.g[food]
t := this.d[p.food]
t.Remove(pair{p.rating, food})
t.Put(pair{newRating, food}, struct{}{})
p.rating = newRating
this.g[food] = p
}
func (this *FoodRatings) HighestRated(cuisine string) string {
return this.d[cuisine].Left().Key.food
}
/**
* Your FoodRatings object will be instantiated and called as such:
* obj := Constructor(foods, cuisines, ratings);
* obj.ChangeRating(food,newRating);
* param_2 := obj.HighestRated(cuisine);
*/
# Accepted solution for LeetCode #2353: Design a Food Rating System
class FoodRatings:
def __init__(self, foods: List[str], cuisines: List[str], ratings: List[int]):
self.d = defaultdict(SortedList)
self.g = {}
for food, cuisine, rating in zip(foods, cuisines, ratings):
self.d[cuisine].add((-rating, food))
self.g[food] = (rating, cuisine)
def changeRating(self, food: str, newRating: int) -> None:
oldRating, cuisine = self.g[food]
self.g[food] = (newRating, cuisine)
self.d[cuisine].remove((-oldRating, food))
self.d[cuisine].add((-newRating, food))
def highestRated(self, cuisine: str) -> str:
return self.d[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)
// Accepted solution for LeetCode #2353: Design a Food Rating System
/**
* [2353] Design a Food Rating System
*
* 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 i^th food,
* cuisines[i] is the type of cuisine of the i^th food, and
* ratings[i] is the initial rating of the i^th 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 * 10^4
* 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] <= 10^8
* 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 * 10^4 calls in total will be made to changeRating and highestRated.
*
*/
pub struct Solution {}
// problem: https://leetcode.com/problems/design-a-food-rating-system/
// discuss: https://leetcode.com/problems/design-a-food-rating-system/discuss/?currentPage=1&orderBy=most_votes&query=
// submission codes start here
struct FoodRatings {}
/**
* `&self` means the method takes an immutable reference.
* If you need a mutable reference, change it to `&mut self` instead.
*/
impl FoodRatings {
fn new(foods: Vec<String>, cuisines: Vec<String>, ratings: Vec<i32>) -> Self {
Self {}
}
fn change_rating(&self, food: String, new_rating: i32) {
//
}
fn highest_rated(&self, cuisine: String) -> String {
String::new()
}
}
/**
* Your FoodRatings object will be instantiated and called as such:
* let obj = FoodRatings::new(foods, cuisines, ratings);
* obj.change_rating(food, newRating);
* let ret_2: String = obj.highest_rated(cuisine);
*/
// submission codes end
#[cfg(test)]
mod tests {
use super::*;
#[test]
#[ignore]
fn test_2353_example_1() {
let mut food_ratings = FoodRatings::new(
vec_string!["kimchi", "miso", "sushi", "moussaka", "ramen", "bulgogi"],
vec_string![
"korean", "japanese", "japanese", "greek", "japanese", "korean",
],
vec![9, 12, 8, 15, 14, 7],
);
assert_eq!(
food_ratings.highest_rated("korean".to_string()),
"kimchi".to_string()
);
assert_eq!(
food_ratings.highest_rated("japanese".to_string()),
"ramen".to_string()
);
food_ratings.change_rating("sushi".to_string(), 16);
assert_eq!(
food_ratings.highest_rated("japanese".to_string()),
"sushi".to_string()
);
food_ratings.change_rating("ramen".to_string(), 16);
assert_eq!(
food_ratings.highest_rated("japanese".to_string()),
"ramen".to_string()
);
}
}
// Accepted solution for LeetCode #2353: Design a Food Rating System
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // 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);
// */
Use this to step through a reusable interview workflow for this problem.
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.
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.
Review these before coding to avoid predictable interview regressions.
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.
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.