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Refactored brackets solution in Clear category for Brackets by likbjorn
def checkio(expression):
expression = ''.join([char for char in expression if char in '[]{}()'])
while True:
l = len(expression)
for o in ['[]', '{}', '()']:
expression = expression.replace(o, '')
if l == len(expression):
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"
Oct. 4, 2016
