Даны два списка, нужно вернуть элементы, которые есть в 1-ом списке, но нет во 2-ом. Оценить эффективность своего решения.

main.py

def symmetric_difference(lst1: list, lst2: list) -> list:
    """Возвращает элементы, которые есть в первом списке, но нет в обоих.

    Скорость работы и используемая дополнительная память - линейна.

    >>> symmetric_difference([], [])
    []
    >>> symmetric_difference([], [1, 2])
    []
    >>> symmetric_difference([1, 2, 3], [])
    [1, 2, 3]
    >>> symmetric_difference([4, 1], [1, 2, 3])
    [4]
    >>> symmetric_difference([1, 2, 3], [2])
    [1, 3]

    """
    make_counter = lambda lst: {el: True for el in lst}
    
    counter1 = make_counter(lst1)
    counter2 = make_counter(lst2)
    
    return [el for el in counter1 if not counter2.get(el)]
    

if __name__ == "__main__":
    import doctest
    doctest.testmod()

'''
Trying:
    symmetric_difference([], [])
Expecting:
    []
ok
Trying:
    symmetric_difference([], [1, 2])
Expecting:
    []
ok
Trying:
    symmetric_difference([1, 2, 3], [])
Expecting:
    [1, 2, 3]
ok
Trying:
    symmetric_difference([4, 1], [1, 2, 3])
Expecting:
    [4]
ok
Trying:
    symmetric_difference([1, 2, 3], [2])
Expecting:
    [1, 3]
ok
Trying:
    symmetric_difference([1, 1, 1, 2, 3], [2])
Expecting:
    [1, 3]
ok

6 passed and 0 failed.
Test passed.
'''