[구름] 1차원 뿌요뿌요 - 파이썬/스택

https://level.goorm.io/exam/49076/1%EC%B0%A8%EC%9B%90-%EB%BF%8C%EC%9A%94%EB%BF%8C%EC%9A%94/quiz/1

내 풀이

from collections import deque

N, M = map(int, input().split())

S = input()
S = list(S)

def bb(S):
    stack = []
    result = []

    S = deque(S)
    if len(set(S)) == 1 and len(S)>1:

        return 'CLEAR!'

    while S:
        elem = S.popleft()
        if stack:
            if stack[-1] == elem:
                stack.append(elem)
                if not S:
                    if len(stack) >= M:
                        return result

            else:
                if len(stack) >= M:
                    result.append(elem)
                    result += list(S)
                    return result

                else:
                    result += stack[:]
                    stack = []
                    stack.append(elem)

        else:
            stack.append(elem)

    result += stack[:]
    return result


while True:
    if S != bb(S):
        if bb(S) != 'CLEAR!':
            S = bb(S)
        else:
            print(bb(S))
            break
    else:
        ans = bb(S)
        print(''.join(ans))
        break

간단한 정답 코드

import sys
input = sys.stdin.readline

N, M = map(int, input().split())
S = input().rstrip()
Q = []
Q.append(('', 1))

S += 'z'
for c in S:
    if Q[-1][0] != c:
        if M <= Q[-1][1]:
            top = Q[-1][0]
            while top == Q[-1][0]:
                Q.pop()
    if Q[-1][0] == c:
        Q.append((c, Q[-1][1] + 1))
    else:
        Q.append((c, 1))
Q.pop()

if len(Q) > 1:
    for c, n in Q:
        print(c, end='')
else:
    print("CLEAR!")

이것저것 조건을 달다보니 코드가 길어졌는데 정답 코드처럼 간소화할 수 있는 문제였다.