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First solution in Clear category for Brackets by mehWincenty
def checkio(expression):
estr = ""
for i in expression :
if i == "(" or i == ")" or i == "[" or i == "]" or i == "{" or i == "}" :
estr += i
estr = list(estr)
pestr = 0
while len(estr) != pestr:
pestr = len(estr)
a = 0
while a < len(estr)-1:
if estr[a]+estr[a+1] == "()":
del estr[a]
del estr[a]
elif estr[a]+estr[a+1] == "[]":
del estr[a]
del estr[a]
elif estr[a]+estr[a+1] == "{}":
del estr[a]
del estr[a]
a += 1
if len(estr) == 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"
Nov. 11, 2016
