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First solution in Clear category for Brackets by Bet
import re
from itertools import cycle
def checkio(expression):
expression = re.sub(r'[^\{\(\[\}\)\]]', "", expression) # カッコ以外の文字を空白に置換
if len(expression) % 2 != 0:
return False
p = '(\(\))|(\[\])|(\{\})' # () [] {} のどれかとマッチするパターン
for bracket in cycle(['()', '[]', '{}']):
expression = expression.replace(bracket, '')
m = re.search(p, expression) # () [] {} のどれも存在しない
if m is None:
return True if expression == '' else False
#These "asserts" using only for self-checking and not necessary for auto-testing
if __name__ == '__main__':
assert checkio("((5+3)*2+1)") == True, "Simple"
assert checkio("{[(3+1)+2]+}") == True, "Different types"
assert checkio("(3+{1-1)}") == False, ") is alone inside {}"
assert checkio("[1+1]+(2*2)-{3/3}") == True, "Different operators"
assert checkio("(({[(((1)-2)+3)-3]/3}-3)") == False, "One is redundant"
assert checkio("2+3") == True, "No brackets, no problem"
Oct. 21, 2019
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