# using a stack data structure
def isValid(s):
if len(s) == 0: return True
if len(s)%2 !=0: return False
l = []
l.append(s[0])
for elem in s[1:]:
if elem == l[-1]:
l.pop() # current element finds a pair in history stack
else:
l.append(elem) # current element does not find a pair in history stack
# the algorithm goes that if an the history is eventually empty, that means all pairs are eliminated
if len(l) == 0:
return True
else:
return False
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