문제
In C#, an IComparer<T> is an interface: any class that implements its Compare method can tell a sort which item comes first. Choosing a different comparer changes the order without changing the sort.
Sort the people by key:
name: by name (ordinal character order, so uppercase letters come before lowercase), and by age when names are equal.age: by age, youngest first, and by name when ages are equal.
Implement sort_by_key(names, ages, key).
Input
names, ages: The people (same length).
key: "name" or "age".
Output
Return the names in sorted order.
제약 조건
0 ≤ number of people ≤ 100,000
0 ≤ age ≤ 150
예시
입력:
key = "age"
ages = [30, 36, 30]
names = ["Lin", "Ada", "Bo"]
예상 출력:
expected = ["Bo", "Lin", "Ada"]입력:
std::string key = "age";
std::vector<long long> ages = {30, 36, 30};
std::vector<std::string> names = {"Lin", "Ada", "Bo"};
예상 출력:
std::vector<std::string> expected = {"Bo", "Lin", "Ada"};입력:
string key = "age";
long[] ages = new long[] {30, 36, 30};
string[] names = new string[] {"Lin", "Ada", "Bo"};
예상 출력:
string[] expected = new string[] {"Bo", "Lin", "Ada"};설명
Bo and Lin are both 30, so they are ordered by name before Ada (36): ["Bo", "Lin", "Ada"].
입력:
key = "name"
ages = [30, 36, 30]
names = ["Lin", "Ada", "Bo"]
예상 출력:
expected = ["Ada", "Bo", "Lin"]입력:
std::string key = "name";
std::vector<long long> ages = {30, 36, 30};
std::vector<std::string> names = {"Lin", "Ada", "Bo"};
예상 출력:
std::vector<std::string> expected = {"Ada", "Bo", "Lin"};입력:
string key = "name";
long[] ages = new long[] {30, 36, 30};
string[] names = new string[] {"Lin", "Ada", "Bo"};
예상 출력:
string[] expected = new string[] {"Ada", "Bo", "Lin"};설명
Alphabetical order: ["Ada", "Bo", "Lin"].
접근 방법
먼저 스스로 풀어 본 뒤 힌트를 하나씩 열어 보세요.
힌트 1
Sort pairs of (name, age) instead of two separate lists.
힌트 2
The chosen key gives the main order; the other field breaks ties.
힌트 3
Use ordinal string comparison so uppercase sorts before lowercase.
예상 복잡도
O(n log n) time, O(n) extra space
참고 자료
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