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First solution in Clear category for Brackets by Mateusz_Tomkowiak
def checkio(expression):
stack = []
pushChars, popChars = "({[", ")}]"
for c in expression:
if c in pushChars:
stack.append(c)
elif c in popChars:
if not len(stack):
return False
else:
stackTop = stack.pop()
balancingBracket = pushChars[popChars.index(c)]
if stackTop != balancingBracket:
return False
else:
continue
return not len(stack)
#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. 23, 2016
