Easy 0 solved 0% acceptance

Unique Records

Remove duplicate records that are equal by value, keeping the first of each in order.

What you'll learn

  • Compare records by their values
  • Remove duplicates while keeping order
  • Normalise fields by trimming spaces

Problem

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.

Constraints

0 ≤ number of records ≤ 100,000

Each field has 1 to 30 characters and no commas

Examples

Example 1
Input:
records = ["Ada,London", "Lin, Hanoi", "Ada , London"]

Expected output:
expected = ["Ada,London", "Lin,Hanoi"]
Input:
std::vector<std::string> records = {"Ada,London", "Lin, Hanoi", "Ada , London"};

Expected output:
std::vector<std::string> expected = {"Ada,London", "Lin,Hanoi"};
Input:
string[] records = new string[] {"Ada,London", "Lin, Hanoi", "Ada , London"};

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

Explanation

After trimming, the third record equals the first, so it is dropped: ["Ada,London", "Lin,Hanoi"].

Example 2
Input:
records = ["ada,Paris", "Ada,Paris"]

Expected output:
expected = ["ada,Paris", "Ada,Paris"]
Input:
std::vector<std::string> records = {"ada,Paris", "Ada,Paris"};

Expected output:
std::vector<std::string> expected = {"ada,Paris", "Ada,Paris"};
Input:
string[] records = new string[] {"ada,Paris", "Ada,Paris"};

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

Explanation

Case matters, so these two records are different: ["ada,Paris", "Ada,Paris"].

How to approach it

Try on your own first, then open one hint at a time.

Hint 1

Trim both fields before comparing records.

Hint 2

A set of (name, city) pairs remembers which records you have seen.

Hint 3

Output each new record the first time it appears, in input order.

Expected complexity

O(total length) time, O(number of records) extra space

References

This challenge is adapted for CodeUtility. Follow the statement and constraints above.

Learn from others

Community solutions

No published solutions yet. Solve this challenge to share the first one.