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Stacky solution in Clear category for Brackets by dronnix
def checkio(expression):
brackets = {'(': ')', '[': ']', '{': '}'}
stack = []
for c in expression:
if c in brackets.keys():
stack.append(c)
if c in brackets.values():
if len(stack) == 0 or c != brackets[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"
Oct. 10, 2014
