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feat: add solutions to lc problem: No.2353 #4114

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169 changes: 141 additions & 28 deletions solution/2300-2399/2353.Design a Food Rating System/README.md
Original file line number Diff line number Diff line change
Expand Up @@ -95,30 +95,42 @@ foodRatings.highestRated("japanese"); // 返回 "ramen"

<!-- solution:start -->

### 方法一
### 方法一:哈希表 + 有序集合

我们可以使用哈希表 $\textit{d}$ 来存储每种烹饪方式下的食物,其中键是烹饪方式,值是一个有序集合,有序集合的每个元素是一个二元组 $(\textit{rating}, \textit{food})$,按照评分从高到低排序,如果评分相同,则按照食物名字的字典序从小到大排序。

我们还可以使用哈希表 $\textit{g}$ 来存储每种食物的评分和烹饪方式。即 $\textit{g}[\textit{food}] = (\textit{rating}, \textit{cuisine})$。

在构造函数中,我们遍历 $\textit{foods}$、$\textit{cuisines}$ 和 $\textit{ratings}$,将每种食物的评分和烹饪方式存储到 $\textit{d}$ 和 $\textit{g}$ 中。

在 $\textit{changeRating}$ 函数中,我们首先获取食物 $\textit{food}$ 的原评分 $\textit{oldRating}$ 和烹饪方式 $\textit{cuisine}$,然后更新 $\textit{g}[\textit{food}]$ 的评分为 $\textit{newRating}$,并从 $\textit{d}[\textit{cuisine}]$ 中删除 $(\textit{oldRating}, \textit{food})$,并将 $(\textit{newRating}, \textit{food})$ 添加到 $\textit{d}[\textit{cuisine}]$ 中。

在 $\textit{highestRated}$ 函数中,我们直接返回 $\textit{d}[\textit{cuisine}]$ 的第一个元素的食物名字即可。

时间复杂度方面,构造函数的时间复杂度为 $O(n \log n)$,其中 $n$ 是食物的数量。其余操作的时间复杂度为 $O(\log n)$。空间复杂度为 $O(n)$。

<!-- tabs:start -->

#### Python3

```python
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 __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:
b, c = self.mp[food]
self.mp[food] = (b, newRating)
self.t[b].remove((c, food))
self.t[b].add((newRating, food))
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.t[cuisine][0][1]
return self.d[cuisine][0][1]


# Your FoodRatings object will be instantiated and called as such:
Expand All @@ -127,36 +139,78 @@ class FoodRatings:
# param_2 = obj.highestRated(cuisine)
```

#### Java

```java
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);
*/
```

#### C++

```cpp
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));
for (int i = 0; i < foods.size(); ++i) {
string food = foods[i], cuisine = cuisines[i];
int rating = ratings[i];
d[cuisine].insert({-rating, food});
g[food] = {rating, cuisine};
}
}

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));
auto [oldRating, cuisine] = g[food];
g[food] = {newRating, cuisine};
d[cuisine].erase({-oldRating, food});
d[cuisine].insert({-newRating, food});
}

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

private:
unordered_map<string, set<pair<int, string>>> d;
unordered_map<string, pair<int, string>> g;
};

/**
Expand All @@ -167,6 +221,65 @@ public:
*/
```

#### Go

```go
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);
*/
```

<!-- tabs:end -->

<!-- solution:end -->
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