[BaekJoon] 3986번 : 좋은 단어 (Python)

코딩 1일 1문제! 오늘의 문제는 백준의 좋은 단어입니다.

2857번: FBI

Solution

def well_word(word_list):
    well_cnt = 0
    
    for word in word_list:
        my_stack = []
        for char in word:
            if my_stack == []:
                my_stack.append(char)
            elif my_stack != []:
                if my_stack[-1] == char:
                    my_stack.pop()
                else:
                    my_stack.append(char)
        if my_stack == []:
            well_cnt += 1
    return well_cnt

if __name__ == "__main__":
    word_list = []
    for _ in range(int(input())):
        word = input()
        word_list.append(word)
        
    print(well_word(word_list))

GitHub - SOMJANG/CODINGTEST_PRACTICE: 1일 1문제 since 2020.02.07