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First solution in Clear category for Brackets by B_dur
def checkio(expression):
stack = []
backets_dict = {"{": "}", "(": ")", "[": "]"}
open_backets = backets_dict.keys()
close_backets = backets_dict.values()
for e in expression:
if e in open_backets:
stack.append(e)
elif e in close_backets:
if not stack:
return False
else:
if e != backets_dict[stack.pop()]:
return False
return len(stack) == 0
#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"
April 17, 2020
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