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Oh my stack solution in Clear category for Brackets by zgub4
def checkio(expression):
stack = list()
for c in expression:
if c == '(':
stack.append(chr(ord(c) + 1))
elif c == '[' or c == '{':
stack.append(chr(ord(c) + 2))
elif c == ')' or c == ']' or c == '}':
if not stack:
return False
else:
if stack[-1] == c:
stack.pop()
else:
return False
return not len(stack)
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("(((([[[{{{3}}}]]]]))))") == False, "Random"
assert checkio("2+3") == True, "No brackets, no problem"
Oct. 21, 2017
