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First solution in Clear category for Brackets by bravebug
def checkio(expression):
brackets = list("({[]})")
just_brackets = ""
for symbol in expression:
if symbol in brackets:
just_brackets += symbol
if len(just_brackets) % 2 != 0:
return False
while len(just_brackets) != 0:
length_before = len(just_brackets)
just_brackets = just_brackets.replace("()", "").replace("{}", "").replace("[]", "")
if len(just_brackets) == length_before:
return False
return True
#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"
Feb. 26, 2020
