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First solution in Clear category for Brackets by botLiker2
def checkio(expression):
stack = []
dict_b = {"{": "}", "[": "]", "(": ")"}
for letter in expression:
if letter in dict_b.keys():
stack.append(letter)
if letter in dict_b.values():
if not (stack and dict_b[stack.pop()] == letter):
return False
return not stack
def test_function():
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"
if __name__ == '__main__':
test_function()
Oct. 30, 2017
