Problem
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.
Constraints
0 ≤ number of people ≤ 100,000
0 ≤ age ≤ 150
Examples
Input:
key = "age"
ages = [30, 36, 30]
names = ["Lin", "Ada", "Bo"]
Expected output:
expected = ["Bo", "Lin", "Ada"]Input:
std::string key = "age";
std::vector<long long> ages = {30, 36, 30};
std::vector<std::string> names = {"Lin", "Ada", "Bo"};
Expected output:
std::vector<std::string> expected = {"Bo", "Lin", "Ada"};Input:
string key = "age";
long[] ages = new long[] {30, 36, 30};
string[] names = new string[] {"Lin", "Ada", "Bo"};
Expected output:
string[] expected = new string[] {"Bo", "Lin", "Ada"};Explanation
Bo and Lin are both 30, so they are ordered by name before Ada (36): ["Bo", "Lin", "Ada"].
Input:
key = "name"
ages = [30, 36, 30]
names = ["Lin", "Ada", "Bo"]
Expected output:
expected = ["Ada", "Bo", "Lin"]Input:
std::string key = "name";
std::vector<long long> ages = {30, 36, 30};
std::vector<std::string> names = {"Lin", "Ada", "Bo"};
Expected output:
std::vector<std::string> expected = {"Ada", "Bo", "Lin"};Input:
string key = "name";
long[] ages = new long[] {30, 36, 30};
string[] names = new string[] {"Lin", "Ada", "Bo"};
Expected output:
string[] expected = new string[] {"Ada", "Bo", "Lin"};Explanation
Alphabetical order: ["Ada", "Bo", "Lin"].
How to approach it
Try on your own first, then open one hint at a time.
Hint 1
Sort pairs of (name, age) instead of two separate lists.
Hint 2
The chosen key gives the main order; the other field breaks ties.
Hint 3
Use ordinal string comparison so uppercase sorts before lowercase.
Expected complexity
O(n log n) time, O(n) extra space
References
This challenge is adapted for CodeUtility. Follow the statement and constraints above.