def solution(s):
    # 특정 괄호 모양이 나오면, 해당하는 것의 반쪽이 반드시 존재해야한다. 
    # 그 상태에서 전체를 돌리면서, 조건에 부합하는 경우 개수를 샌다 
    # target 은 enumerate 되면서 복사 된듯 
    target = s 
    answer = 0    
    loop = 0
    # 1차 loop 로 감쌈. 
    # 2차 전체 검사 및 문제 없을시 answer += 1
    # while loop < len(s):
    #     answer += checker(target)
    #     target = target[1:] + target[0]
    #     loop += 1
    for i in range(len(s)):
        answer += checker(target)
        target = target[1:] + target[0]
    return answer

def checker(string):
    origin = "{[("
    opposite = "}])"
    for i, word in enumerate(string):
        if word in origin:
            if opposite[origin.index(word)] not in string[i+1:]:
                return 0
            else:
                for j in range(len(string)):
                    j += i
                    if string[j] == opposite[origin.index(word)]:
                        if (j - i - 1) % 2 != 0:
                            return 0
                        break
        else:
            if origin[opposite.index(word)] not in string[:i]:
                return 0
    return 1

# 해결 방법...! 
def solution(s):
    answer = 0
    
    for _ in range(len(s)): 
        if checkBracket(s):
            answer += 1
        s = s[1:] + s[0]
        
    return answer

def checkBracket(target):
    stack = []
    bracketMap = {'}': '{', ']': '[', ')': '('}
    
    for char in target:
        if char in bracketMap.values():
            stack.append(char)
        else :
            if stack:
                if stack[-1] == bracketMap[char]:
                    stack.pop()
                else:
                    return False
            else:
                return False
    return not stack

from collections import deque

def check(s):
    while True:
        if "()" in s: s=s.replace("()","")
        elif "{}" in s: s=s.replace("{}","")
        elif "[]" in s: s=s.replace("[]","")
        else: return False if s else True       

def solution(s):
    ans = 0
    que = deque(s)

    for i in range(len(s)):
        if check(''.join(que)): ans+=1
        que.rotate(-1)
    return ans