Regular Bracket Sequence (Codeforces 1469A)
Tags: Greedy, constructive algorithms

A bracket sequence is called regular if it is possible to obtain correct arithmetic expression by inserting characters + and 1 into this sequence. For example, sequences (())(), () and (()(())) are regular, while )(, (() and (()))( are not. Let's call a regular bracket sequence "RBS".You are given a sequence $$$s$$$ of $$$n$$$ characters (, ), and/or ?. There is exactly one character ( and exactly one character ) in this sequence.You have to replace every character ? with either ) or ( (different characters ? can be replaced with different brackets). You cannot reorder the characters, remove them, insert other characters, and each ? must be replaced.Determine if it is possible to obtain an RBS after these replacements.

Input: The first line contains one integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases.Each test case consists of one line containing $$$s$$$ ($$$2 \le |s| \le 100$$$) — a sequence of characters (, ), and/or ?. There is exactly one character ( and exactly one character ) in this sequence.

Output: For each test case, print YES if it is possible to obtain a regular bracket sequence, or NO otherwise}. You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).

Note: In the first test case, the sequence is already an RBS.In the third test case, you can obtain an RBS as follows: ()() or (()).In the fourth test case, you can obtain an RBS as follows: ()().

Sample Inputs:
5
()
(?)
(??)
??()
)?(?

Sample Outputs:
YES
NO
YES
YES
NO

Solution (Python):
t = int(input())
while t > 0:
    s = input()
    if len(s) % 2 == 0 and s[0] != ')' and (s[len(s) - 1] != '('):
        print('YES')
    else:
        print('NO')
    t -= 1

Solution (Python):
t = int(input())
while t > 0:
    t -= 1
    s = input()
    if len(s) % 2 == 1:
        print('NO')
    elif s[0] != ')' and s[-1] != '(':
        print('YES')
    else:
        print('NO')

Solution (Python):
for _ in range(int(input())):
    s = input()
    if len(s) % 2 != 1 and s[0] != ')' and (s[-1] != '('):
        print('Yes')
        continue
    print('No')