Problema
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.
Restricciones
0 ≤ number of people ≤ 100,000
0 ≤ age ≤ 150
Ejemplos
Entrada:
key = "age"
ages = [30, 36, 30]
names = ["Lin", "Ada", "Bo"]
Salida esperada:
expected = ["Bo", "Lin", "Ada"]Entrada:
std::string key = "age";
std::vector<long long> ages = {30, 36, 30};
std::vector<std::string> names = {"Lin", "Ada", "Bo"};
Salida esperada:
std::vector<std::string> expected = {"Bo", "Lin", "Ada"};Entrada:
string key = "age";
long[] ages = new long[] {30, 36, 30};
string[] names = new string[] {"Lin", "Ada", "Bo"};
Salida esperada:
string[] expected = new string[] {"Bo", "Lin", "Ada"};Explicación
Bo and Lin are both 30, so they are ordered by name before Ada (36): ["Bo", "Lin", "Ada"].
Entrada:
key = "name"
ages = [30, 36, 30]
names = ["Lin", "Ada", "Bo"]
Salida esperada:
expected = ["Ada", "Bo", "Lin"]Entrada:
std::string key = "name";
std::vector<long long> ages = {30, 36, 30};
std::vector<std::string> names = {"Lin", "Ada", "Bo"};
Salida esperada:
std::vector<std::string> expected = {"Ada", "Bo", "Lin"};Entrada:
string key = "name";
long[] ages = new long[] {30, 36, 30};
string[] names = new string[] {"Lin", "Ada", "Bo"};
Salida esperada:
string[] expected = new string[] {"Ada", "Bo", "Lin"};Explicación
Alphabetical order: ["Ada", "Bo", "Lin"].
Cómo abordarlo
Inténtalo primero por tu cuenta y luego abre las pistas una a una.
Pista 1
Sort pairs of (name, age) instead of two separate lists.
Pista 2
The chosen key gives the main order; the other field breaks ties.
Pista 3
Use ordinal string comparison so uppercase sorts before lowercase.
Complejidad esperada
O(n log n) time, O(n) extra space
Referencias
Este reto está adaptado para CodeUtility. Sigue el enunciado y las restricciones indicadas arriba.