Create a complement of list preserving duplicate values

Question

Given list a = [1, 2, 2, 3] and its sublist b = [1, 2] find a list complementing b in such a way that sorted(a) == sorted(b + complement)

Answer 1

Main idea: if the values are not unique, make them unique

def add_duplicate_position(items):
    element_counter = {}
    for item in items:
        element_counter[item] = element_counter.setdefault(item,-1) + 1
        yield element_counter[item], item
    
assert list(add_duplicate_position([1, 2, 2, 3])) == [(0, 1), (0, 2), (1, 2), (0, 3)]

def create_complementary_list_with_duplicates(a,b):
    a = list(add_duplicate_position(a))
    b = set(add_duplicate_position(b))
    return [item for _,item in [x for x in a if x not in b]]
  
a = [1, 2, 2, 3]
b = [1, 2]
assert create_complementary_list_with_duplicates(a,b) == [2, 3]