They shouldn't define the assignment as "debugging", it's misleading because "debugging" means a very specific process which requires test or application code with a use case. What they meant to say was "notice a bug", or "there is a bug in this code, what is it?", or "Is there bug in this code?". Also, there was no code on the screen at all. ...After a few next minutes looking for code and promised "provided test files", and pushing randomly buttons I got a message saying that the "test files" cost 5 hackos. Wouldn't be better at the begining of the assignment to inform the reader where the "test files" are and that they would cost money, and also from the beggining on to state that one should open a paying account?

class EvenStream: def init(self): self.current = 0

def get_next(self):
    next_value = self.current
    self.current += 2
    return next_value

class OddStream: def init(self): self.current = 1

def get_next(self):
    next_value = self.current
    self.current += 2
    return next_value

def print_from_stream(n, stream=None): if stream is None: stream = EvenStream() for _ in range(n): print(stream.get_next())

if name == "main": queries = int(input()) for _ in range(queries): stream_name, n = input().split() n = int(n) if stream_name == "even": print_from_stream(n) elif stream_name == "odd": print_from_stream(n, OddStream())

with no code given, im wondering where to debug

class EvenStream(object): def init(self): self.current = 0

def get_next(self):
    to_return = self.current
    self.current += 2
    return to_return

class OddStream(object): def init(self): self.current = 1

def get_next(self):
    to_return = self.current
    self.current += 2
    return to_return

def print_from_stream(n, stream=EvenStream()): stream.init() for _ in range(n): print(stream.get_next())

queries = int(input()) for _ in range(queries): stream_name, n = input().split() n = int(n) if stream_name == "even": print_from_stream(n) else: print_from_stream(n, OddStream())