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First solution in Clear category for Brackets by testarossa
def checkio(expression):
stack = []
for c in expression:
if c in set(["{", "[", "("]):
stack.append(c)
if c == "}":
if len(stack) <= 0 or stack[-1] != "{":
return False
stack.pop()
if c == "]":
if len(stack) <= 0 or stack[-1] != "[":
return False
stack.pop()
if c == ")":
if len(stack) <= 0 or stack[-1] != "(":
return False
stack.pop()
if len(stack) > 0:
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. 19, 2016
