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stack solution in Clear category for Brackets by Krischtopp
def checkio(expression):
stack = []
for char in expression:
try:
if char in ("(", "[", "{"):
stack.append(char) #Remember open brackets
elif char == ")":
if "(" != stack.pop():
return False #Closed bracket doesn't match the open bracket
elif char == "]":
if "[" != stack.pop():
return False #Closed bracket doesn't match the open bracket
elif char == "}":
if "{" != stack.pop():
return False #Closed bracket doesn't match the open bracket
except IndexError:
return False #More closed brackets than open brackets
return not stack #There need to be no open brackets left
#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"
Jan. 15, 2016
