my solution: import re for i in range(int(input())): try: re.compile(input()) print(True) except re.error: print(False) but it is giving output as TRUE TRUE and hence failing the sample test case. Although i did solve it using python 2 can anyone help me with a pypy3 solution

This works for Python 3 import re if name == 'main': n = int(raw_input()) for i in range(n): try: re.compile(raw_input()) print("True") except re.error: print("False")

true true this code is Als gave me

Explanation: sys.stdin.readline() reads raw input lines safely.

.strip() removes the newline character at the end.

We compile the pattern using re.compile().

If it's a valid regex, True is printed; otherwise, False.

import re import sys

n = int(sys.stdin.readline()) for _ in range(n): pattern = sys.stdin.readline().strip() try: re.compile(pattern) print(True) except re.error: print(False)

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)