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using stack solution in Clear category for Brackets by cheikhnamouna
def checkio(e):
s, opened, closed = [], '{[(', '}])'
for c in e:
if c in opened:
s.append(c)
elif c in closed:
if not s or s.pop() != opened[closed.find(c)]:
return False
return not s
Dec. 30, 2018
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