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First solution in Clear category for Brackets by wojtekpaciej
def checkio(data):
# replace this for solution
left_stack = []
for i in data:
if i == '(' or i == '[' or i == '{':
left_stack.append(i)
elif i == ")":
if len(left_stack) == 0 or left_stack[-1] != '(':
return False
else:
left_stack.pop()
elif i == ']':
if len(left_stack) == 0 or left_stack[-1] != '[':
return False
else:
left_stack.pop()
elif i == '}':
if len(left_stack) == 0 or left_stack[-1] != '{':
return False
else:
left_stack.pop()
if len(left_stack) == 0:
return True
else:
return False
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"
Dec. 12, 2016
