Why doesn't != work for string index comparison?

I tried to run this code with input "AC039"

 code = input("Enter code: ").upper()
 if code[0] != ('N' or 'A' or 'C' ):
     print("The first character must be N, A or C")
else:
    print("Pass!")

It gave me the output error result:

The first character must be N, A or C

However, if I input "AC039" into the below code using ‘not in’,

code = input("Enter code: ").upper()
if code[0] not in ["N", "A", "C"]:
    print("The first character must be N, A or C")
else:
    print("Pass!")

The resulting output is:

print("Pass!")

Why doesn’t "!=" work for the first set of code, since both code[0] and ‘A’ are string types?

I ran a check using type function on code[0] and it returned string type.

code = input("Enter code: ").upper()
print(type(code[0]))
print(type('A'))

returns:

<class 'str'>
<class 'str'>

>Solution :

It’s because the "or" operator compares sequentially.
from the given input "AC039",

code[0] != ('N' or 'A' or 'C' ):

what’s happening from the above code is, comparing one by one.

(code[0] != 'N') -> True
(code[0] != 'A') -> False
(code[0] != 'C') -> True

since the first ‘N’ doesn’t match with ‘A’, it stops going through other comparisons, ‘A’ and ‘C’.

print('N' != ('N' or 'A' or 'C' )) -> False
print('A' != ('N' or 'A' or 'C' )) -> True
print('C' != ('N' or 'A' or 'C' )) -> True

However, if code[0] not in ["N", "A", "C"]: works because "not in" checks whole elements in the array.

I hope it helps.