Unique Records

C# records compare by value: two records are equal when all their fields are equal, even if they are different objects. That makes removing duplicates easy.

Each record is written name,city. Trim spaces around both fields; after trimming, two records are equal when both fields match exactly (case matters).

Implement unique_records(records).

Input

records: The records, in order.

Output

Return the distinct records in the order they first appear, written name,city with the trimmed fields.

เงื่อนไข

0 ≤ number of records ≤ 100,000

Each field has 1 to 30 characters and no commas

ตัวอย่าง

อินพุต:
records = ["Ada,London", "Lin, Hanoi", "Ada , London"]

ผลลัพธ์ที่คาดหวัง:
expected = ["Ada,London", "Lin,Hanoi"]

อินพุต:
std::vector<std::string> records = {"Ada,London", "Lin, Hanoi", "Ada , London"};

ผลลัพธ์ที่คาดหวัง:
std::vector<std::string> expected = {"Ada,London", "Lin,Hanoi"};

อินพุต:
string[] records = new string[] {"Ada,London", "Lin, Hanoi", "Ada , London"};

ผลลัพธ์ที่คาดหวัง:
string[] expected = new string[] {"Ada,London", "Lin,Hanoi"};

อินพุต:
records = ["ada,Paris", "Ada,Paris"]

ผลลัพธ์ที่คาดหวัง:
expected = ["ada,Paris", "Ada,Paris"]

อินพุต:
std::vector<std::string> records = {"ada,Paris", "Ada,Paris"};

ผลลัพธ์ที่คาดหวัง:
std::vector<std::string> expected = {"ada,Paris", "Ada,Paris"};

อินพุต:
string[] records = new string[] {"ada,Paris", "Ada,Paris"};

ผลลัพธ์ที่คาดหวัง:
string[] expected = new string[] {"ada,Paris", "Ada,Paris"};

แหล่งอ้างอิง

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• Microsoft Learn, Records