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Second solution in Uncategorized category for Brackets by Elena_Korljukova
def checkio(expression):
a = ''.join(list(filter(lambda x: x in '[]{}()', expression)))
while any(i in a for i in ['[]','{}', '()']):
for x in ['[]','{}', '()']:
a = a.replace(x, '')
return len(a) == 0
#These "asserts" using only for self-checking and not necessary for auto-testing
if __name__ == '__main__':
print(checkio("(({[(((1)-2)+3)-3]/3}-3)"))
#print(checkio("(({[(((1)-2)+3)-3]/3}-3)"))
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"
Aug. 20, 2020
