[프로그래머스/Programmers] 괄호 변환 - Python

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코딩테스트 연습 - 괄호 변환

 

Solution

def isBalanced(s):
    return s.count('(') == s.count(')')

def isCorrect(s):
    if s[0] == ')':
        return False

    count = 0
    for i in range(len(s)):
        if s[i] == '(':
            count += 1
        else:
            count -= 1

        if count < 0:
            return False

    return count == 0

def solution(p):
    answer = ''
    u = ''
    v = ''

    if len(p) == 0 or isCorrect(p):
        return p

    for i in range(2, len(p) + 1, 2):
        if isBalanced(p[0:i]):
            u = p[0:i]
            v = p[i:]
            break

    if isCorrect(u):
        answer += u + solution(v)
    else:
        answer += '(' + solution(v) + ')'
        for c in u[1:-1]:
            if c == '(':
                answer += ')'
            else:
                answer += '('

    return answer


    return p

 

Other Solution

def solution(p):
    if p == '':
        return p

    chk = True
    count = 0
    for i in range(len(p)):
        if p[i] == '(':
            count += 1
        else:
            count -= 1

        if count < 0:
            chk = False

        if count == 0:
            if chk:
                return p[:i+1] + solution(p[i + 1:])
            else:
                return '(' + solution(p[i + 1:]) + ')' + ''.join(list(map(lambda x: '(' if x == ')' else ')', p[1:i])))