Problema
For each integer from 1 through limit, use Fizz for multiples of 3, Buzz for multiples of 5, and FizzBuzz for multiples of both. Otherwise use the number written as text. Return the entries in order instead of printing them.
Implement fizz_buzz(limit).
Input
limit: The last integer to include.
Output
Return a list of strings, one for each integer from 1 through limit.
Restricciones
0 ≤ limit ≤ 10,000. A limit of 0 produces an empty list.
Ejemplos
Entrada:
limit = 5
Salida esperada:
expected = ["1", "2", "Fizz", "4", "Buzz"]Entrada:
long long limit = 5;
Salida esperada:
std::vector<std::string> expected = {"1", "2", "Fizz", "4", "Buzz"};Entrada:
long limit = 5;
Salida esperada:
string[] expected = new string[] {"1", "2", "Fizz", "4", "Buzz"};Explicación
3 is divisible by 3 and 5 is divisible by 5. Other entries keep their number.
Entrada:
limit = 0
Salida esperada:
expected = []Entrada:
long long limit = 0;
Salida esperada:
std::vector<std::string> expected = {};Entrada:
long limit = 0;
Salida esperada:
string[] expected = new string[] {};Explicación
There are no integers from 1 through 0, so the list is empty.
Cómo abordarlo
Inténtalo primero por tu cuenta y luego abre las pistas una a una.
Pista 1
Each position depends only on the current integer’s divisibility by 3 and 5.
Pista 2
Build the result list in order by checking both divisibility conditions for every number.
Pista 3
Handle the lower bound carefully: if the limit is 0, the answer is an empty list.
Complejidad esperada
O(n) time, O(n) extra space
Referencias
Este reto está adaptado para CodeUtility. Sigue el enunciado y las restricciones indicadas arriba.