python 3 solution

python added support for possesive quantifiers (*+,++,?+,{ }+) from python 3.11, but the expected outputs are not updated to align with the updated python version. so we need to handle possesive quantifiers separately.

import re
res="False"
pattern = r'(?:[*+?]|\{\d+(?:,\d*)?\})\+'    # ++,*+,?+,{..}+
for _ in range(int(input())):
    inp=input()
    if re.findall(pattern,inp):
        res="False"
    else:
        try:
            re.compile(inp)
            res="True"
        except re.error as e:
            res="False"
    
    print(res)

I solve this problem with Python 2 and raw_string. But in Python 3, I don't know how it's possible cause we always have the SyntaxError with '\'.

Nevertheless, here my code :

import re

T = int(input())

for T_itr in range(T):
    regex = input()

    try:
        re.compile(regex)
        print("True")
    except re.error:
        print("False")
        

Try Like this

import re

T = int(input()) invalid_repeats = ['++', '**', '??', '+', '+', '?', '?', '+?', '??']

for _ in range(T): S = input().strip()

# Basic invalid pattern check
is_invalid = any(rep in S for rep in invalid_repeats)

if is_invalid:
    print("False")
    continue

try:
    re.compile(S)
    print("True")
except re.error:
    print("False")

how to solve this code

Accepted solution for python 2

import re
for _ in range(int(raw_input())):
    try:
        re.compile(raw_input().strip())
        print(True)
    except re.error:
        print(False)