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simple stack solution in Clear category for Brackets by kalauroma7997
def checkio(expression):
brackets = {'{': '}', '[': ']', '(': ')'}
stack = []
for a in expression:
if a in brackets.keys():
stack.append(brackets[a])
elif a in brackets.values():
if not stack: return False
elif stack.pop() != a: 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"
Nov. 9, 2020
