[Mock] Google 2

1021. Remove Outermost Parentheses

A valid parentheses string is either empty (""), "(" + A + ")", or A + B, where A and B are valid parentheses strings, and + represents string concatenation. For example, "", "()", "(())()", and "(()(()))" are all valid parentheses strings.

A valid parentheses string S is primitive if it is nonempty, and there does not exist a way to split it into S = A+B, with A and B nonempty valid parentheses strings.

Given a valid parentheses string S, consider its primitive decomposition: S = P_1 + P_2 + ... + P_k, where P_i are primitive valid parentheses strings.

Return S after removing the outermost parentheses of every primitive string in the primitive decomposition of S.

My Answer 1: Accepted (Runtime: 28 ms - 98.07% / Memory Usage: 14.3 MB - 86.08%)

class Solution:
    def removeOuterParentheses(self, S: str) -> str:
        temp = ""
        result = ""
        isOuter = False
        count = 0
        
        for i in range(len(S)):
            # 시작 괄호는 isOuter
            if not isOuter and S[i] == "(":
                isOuter = True
                count = 0
                continue
            
            # Outer 가 있는 상태면 모두 temp 에 넣어주기
            if isOuter:
                if S[i] == "(":
                    temp += "("
                    count += 1
                else:
                    # count 가 0 인데 ")" 를 또 만나면 Outer 짝꿍이니까 result 에 update
                    if count == 0:
                        result += temp
                        isOuter = False
                        temp = ""
                    else:
                        temp += ")"
                        count -= 1
        
        return result

바깥 괄호를 만나면 isOuter = True => 짝꿍 ) 를 만날 때까지 temp 에 넣어주면서 괄호 count

if count == 0: 인데 ) 를 만났으면 아까 못 찾은 짝꿍이니까 result 에 update

어려운게 아닌데 머리가 안돌아가서 좀 걸렸네요;

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1025. Divisor Game

Alice and Bob take turns playing a game, with Alice starting first.

Initially, there is a number N on the chalkboard. On each player's turn, that player makes a move consisting of:

• Choosing any x with 0 < x < N and N % x == 0.
• Replacing the number N on the chalkboard with N - x.

Also, if a player cannot make a move, they lose the game.

Return True if and only if Alice wins the game, assuming both players play optimally.

My Answer 1: Accepted (Runtime: 28 ms - 85.49% / Memory Usage: 14.1 MB - 72.25%)

class Solution:
    def divisorGame(self, N: int) -> bool:
        # 0 이면 Bob 우승 => False, 1 이면 Alice 우승 => True
        winner = 0
        
        if N == 1:
            return False
        
        i = N-1
        
        while N != i and N and i != 0:
            if N % i == 0:
                N = N - i
                winner = not winner
            else:
                i -= 1
        
        if winner:
            return True
        return False

while 문 돌려서 N 의 약수를 찾으면 winner 바꿔주고 N 은 N - 약수로 update

최종적으로 winner 가 0 이면 Bob 우승 => False, 1 이면 Alice 우승 => True