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Brackets solution in Clear category for Brackets by graffme
def checkio(expression):
#Zrób tablicę z wszystkimi bracketami rozpoczynającymi
brackets = []
result = True
for char in expression:
if char in ["{", "[", "("]:
brackets.append(char)
#jeśli napotkasz bracket zamykający i w brackets coś jest, sprawdź czy ostatni element jest taki sam jak napotkany, jak tak to go wywal, jak nie to jest źle!
if char == ")":
if len(brackets) <= 0 or brackets[-1] != "(":
result = False
elif brackets[-1] == "(":
brackets.pop()
elif char == "}":
if len(brackets) <= 0 or brackets[-1] != "{":
result = False
elif brackets[-1] == "{":
brackets.pop()
elif char == "]":
if len(brackets) <= 0 or brackets[-1] != "[":
result = False
elif brackets[-1] == "[":
brackets.pop()
if len(brackets) != 0:
result = False
return result
#These "asserts" using only for self-checking and not necessary for auto-testing
if __name__ == '__main__':
assert checkio("[[[1+[1+1]]])") == False, "No brackets, no problem"
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. 12, 2017
