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Sub and replace solution in Clear category for Brackets by MuxaJlbl4
from re import sub
def checkio(expression):
brackets = ['()','[]','{}']
expression = sub('[^(){}\[\]]?','',expression)
# Remove all single pairs, while its available
while any(brackets_pair in expression for brackets_pair in brackets):
for brackets_pair in brackets:
expression = expression.replace(brackets_pair, '')
return 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"
Dec. 1, 2020