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First solution in Clear category for Brackets by u2ezi
OPEN_BRACKETS = ("{","[","(")
CLOSE_BRACKETS = ("}","]",")")
def checkio(expression):
# let's have a stack of "open brackets"
stack = []
# parse the expression, extract the brackets
for ch in expression:
if ch in OPEN_BRACKETS:
# open bracket goes into the stack
stack.append(OPEN_BRACKETS[OPEN_BRACKETS.index(ch)])
if ch in CLOSE_BRACKETS:
# if stack is empty it means we have an orphan closing bracket
if len(stack) == 0:
return False
# closed bracket it compared to the top of the stack
if stack.pop() != OPEN_BRACKETS[CLOSE_BRACKETS.index(ch)]:
# this means we didn't close them in order
return False
# if there's anything left in the stack it means that we have orphan brackets
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. 31, 2016
