FizzBuzz (el clásico de las entrevistas).

// Challenge: FizzBuzz (the interview classic).
// Print the numbers from 1 to n, but:
// - If it's a multiple of 3, print "Fizz".
// - If it's a multiple of 5, print "Buzz".
// - If it's a multiple of both 3 and 5, print "FizzBuzz".
// Input: n = 15
// Expected output: 1, 2, Fizz, 4, Buzz, Fizz, 7, 8, Fizz, Buzz, 11, Fizz, 13, 14, FizzBuzz

// Solution: we iterate from 1 to n and decide what to add.
// Order matters: we first check the case of both 3 AND 5.
List<String> fizzbuzz(int n) {
  final result = <String>[];
  for (var i = 1; i <= n; i++) {
    if (i % 3 == 0 && i % 5 == 0) {
      result.add('FizzBuzz');
    } else if (i % 3 == 0) {
      result.add('Fizz');
    } else if (i % 5 == 0) {
      result.add('Buzz');
    } else {
      result.add('$i');
    }
  }
  return result;
}

void main() {
  print(fizzbuzz(15).join(', '));
  // 1, 2, Fizz, 4, Buzz, Fizz, 7, 8, Fizz, Buzz, 11, Fizz, 13, 14, FizzBuzz

  print(fizzbuzz(5).join(', '));
  // 1, 2, Fizz, 4, Buzz

  // Explanation:
  // The % (modulo) operator gives the remainder of a division. If i % 3 is 0,
  // the number is a multiple of 3. The key is to check FIRST the case of a
  // multiple of both 3 and 5; if we left it for the end, it would never be
  // reached because the earlier conditions would catch it first.
}