Question

def clean_up(l):
'''list of str->list of str

The functions takes as input a list of characters.
It returns a new list containing the same characters as l except that
one of each characters that appears odd number of times in l is removed
and all the * characters are removed

>>> clean_up(['A', '*', '$', 'C', '*', '*', 'P', 'E', 'D', 'D', '#', 'D', 'E', 'B', '$', '#'])
['#', '#', '$', '$', 'D', 'D', 'E', 'E']

>>> clean_up(['A', 'B', '*', 'C', '*', 'D', '*', '*', '*', 'E'])
[]
'''

Answer 1

def clean_up(l):
    '''list of str->list of str

    The functions takes as input a list of characters.
    It returns a new list containing the same characters as l except that
    one of each characters that appears odd number of times in l is removed
    and all the * characters are removed
    '''
    result = []
    for x in l:
        if (x!='*') and (l.count(x)%2==0):
            result.append(x)
    return result

# Testing
print(clean_up(['A', '*', '$', 'C', '*', '*', 'P', 'E', 'D', 'D', '#', 'D', 'E', 'B', '$', '#']))
print(clean_up(['A', 'B', '*', 'C', '*', 'D', '*', '*', '*', 'E']))