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First - Brackets solution in Clear category for Brackets by denys.contant
def checkio(expression):
stack = []
brackets = {"(":')', '[':']', '{':'}', }
opening_brackets = brackets.keys()
closing_brackets = brackets.values()
for char in expression:
if char in opening_brackets:
stack.append(char)
if char in closing_brackets:
if len(stack)==0 or char != brackets[stack.pop()]:
return False
return stack == []
Oct. 9, 2018