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ord solution in Creative category for Brackets by nkapliev
def checkio(expression):
stack = []
for char in expression:
if char in '[{(':
stack.append(char)
elif char in ']})':
if not stack or abs(ord(char) - ord(stack.pop())) > 2:
return False
return not 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"
March 3, 2015
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