Enable Javascript in your browser and then refresh this page, for a much enhanced experience.
First solution in Clear category for Brackets by Andreas_Strus
def checkio(expression):
exp = expression[:]
bra = ['[', ']', '(', ")", '{', '}']
for n in ['0', '1', '2', '3', '4', '5', '6', '7', '8', '9', '+', '-', '*', '/', '=']:
if n not in bra:
exp = exp.replace(n, "")
d = 0
e = len(exp)
for d in range(len(exp)):
exp = exp.replace("()", "")
exp = exp.replace("[]", "")
exp = exp.replace("{}", "")
if exp != "":
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"
Nov. 9, 2016
