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First solution in Clear category for Brackets by manul4eg
openbrackets = ["(","[","{"]
closebrackets = [")","]","}",]
def checkio(expression):
temp = []
for i in expression:
if i in openbrackets:
temp.append(openbrackets[openbrackets.index(i)])
if i in closebrackets:
if len(temp) == 0:
return False
if temp.pop() != openbrackets[closebrackets.index(i)]:
return False
return len(temp) == 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"
Oct. 7, 2016
