Enable Javascript in your browser and then refresh this page, for a much enhanced experience.
First solution in Clear category for Brackets by Bartosz_M
brackets = ['()', '{}', '[]', '<>']
brackets_left = {a for a, _ in brackets}
brackets_right = {b: a for a, b in brackets}
def checkio(expression):
expression_brackets = []
for i in expression:
if i in brackets_left:
expression_brackets.append(i)
elif i in brackets_right:
if not expression_brackets or expression_brackets[-1] != brackets_right[i]:
return False
expression_brackets.pop()
return not expression_brackets
#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. 22, 2016
