Cancelling out duplicates in a list by even parity

Question

Let's say we have a list such that list = ["1", "2", "1", "3", "3", "4", "3"]

I need a way to remove any duplicates and cancel them out in case there is a total even number of those duplicates. Otherwise, if there is an odd number, only one element would remain. For example, our list would filter to ["2", "3", "4"]

In my case, this list would initially be empty elements added during runtime. I need little time complexity as this list is used in real time for drawing shapes (trying my luck in pygame).

I came across this potential solution to my problem:

def my_function(x):
  return list(dict.fromkeys(x))

mylist = my_function(["a", "b", "a", "c", "c"])
print(mylist) #from https://www.w3schools.com/python/python_howto_remove_duplicates.asp

I can't get my head around how to implement this cancelling out without overcomplicating it. Thank you for reading.

Answer 1

Using Counter..!

Counter - A subclass of dict that's specially designed for counting hashable objects in Python / occurences of the objects.

Code: Time-O(n) Space-O(n)

from collections import Counter
lis = ["1", "2", "1", "3", "1" , "3", "4", "3"]

a=Counter(lis)
res=[]
for key,value in a.items():
    if value%2!=0:
        res.append(key)
print(res)

Output:

['1', '2', '3', '4']

I think you made a typo in first example output.. 1 should also added cause 1 occurence is odd..

Below code by dictionary..! Time- O(n) Space-O(n)

Code:

lis = ["1", "2", "1", "3", "1" , "3", "4", "3"]
dic,res={},[]

for num in lis:
    dic[num]=dic.get(num,0)+1
    
for key,value in dic.items():
    if value%2!=0:
        res.append(key)
print(res)  #Same output.

Updated part.

list comprehension.

lis = ["1", "2", "1", "3", "1" , "3", "4", "3"]
dic={}

for num in lis:
    dic[num]=dic.get(num,0)+1
    
res=[key for key,value in dic.items() if value%2!=0]

print(res)  #Same output.