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First solution in Clear category for Brackets by checinski.szymon
def compareBrackets(b1, b2):
if b1 == ")":
return b2 == "("
if b1 == "]":
return b2 == "["
if b1 == "}":
return b2 == "{"
def checkio(expression):
i = 0
list = []
while i < len(expression):
if expression[i] == "(" or expression[i] == "[" or expression[i] == "{":
list.append(expression[i])
elif expression[i] == ")" or expression[i] == "]" or expression[i] == "}":
if len(list) == 0:
return False
if not compareBrackets(expression[i], list[len(list) - 1]):
return False
else:
if len(list) > 0:
list.pop()
i += 1
if len(list) > 0:
print()
return False
else:
return True
#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. 6, 2016
