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First solution in Clear category for Brackets by lemaachi
def checkio(expression):
ouvr = ['(','[','{']
ferm = [')',']','}']
pile = []
for i in expression :
if i in ouvr :
pile.append(ouvr.index(i))
elif i in ferm :
if len(pile) == 0 :
return False
elif ferm.index(i) != pile[len(pile)-1] :
return False
else :
pile.pop()
if len(pile) == 0 :
return True
else :
return False
#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"
April 29, 2020
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