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re.sub solution in Clear category for Brackets by jtokaz
import re
def checkio(expression):
expression = re.sub(r'[^(){}\[\]]','',expression)
for _ in range(len(expression)//2):
expression = re.sub(r'\(\)|\[\]|\{\}','',expression)
return not bool(expression)
#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"
June 20, 2019
