Problema
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.
Vincoli
0 ≤ number of records ≤ 100,000
Each field has 1 to 30 characters and no commas
Esempi
Input:
records = ["Ada,London", "Lin, Hanoi", "Ada , London"]
Output atteso:
expected = ["Ada,London", "Lin,Hanoi"]Input:
std::vector<std::string> records = {"Ada,London", "Lin, Hanoi", "Ada , London"};
Output atteso:
std::vector<std::string> expected = {"Ada,London", "Lin,Hanoi"};Input:
string[] records = new string[] {"Ada,London", "Lin, Hanoi", "Ada , London"};
Output atteso:
string[] expected = new string[] {"Ada,London", "Lin,Hanoi"};Spiegazione
After trimming, the third record equals the first, so it is dropped: ["Ada,London", "Lin,Hanoi"].
Input:
records = ["ada,Paris", "Ada,Paris"]
Output atteso:
expected = ["ada,Paris", "Ada,Paris"]Input:
std::vector<std::string> records = {"ada,Paris", "Ada,Paris"};
Output atteso:
std::vector<std::string> expected = {"ada,Paris", "Ada,Paris"};Input:
string[] records = new string[] {"ada,Paris", "Ada,Paris"};
Output atteso:
string[] expected = new string[] {"ada,Paris", "Ada,Paris"};Spiegazione
Case matters, so these two records are different: ["ada,Paris", "Ada,Paris"].
Come affrontarlo
Prova prima da solo, poi apri un suggerimento alla volta.
Suggerimento 1
Trim both fields before comparing records.
Suggerimento 2
A set of (name, city) pairs remembers which records you have seen.
Suggerimento 3
Output each new record the first time it appears, in input order.
Complessità attesa
O(total length) time, O(number of records) extra space
Riferimenti
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