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First solution in Clear category for Brackets by lblc
def checkio(expression):
br = "(){}[]"
x = ""
for l in expression:
if br.find(l) >= 0:
x += l
while (x.find("()") >= 0) or (x.find("[]") >= 0) or (x.find("{}") >= 0):
x = x.replace("()", "")
x = x.replace("[]", "")
x = x.replace("{}", "")
return len(x) == 0
#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"
Oct. 18, 2016
