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First solution in Clear category for Brackets by Krzysztof_Dziedzic
OPEN = ("{","[","(")
CLOSE = ("}","]",")")
def checkio(expression):
st = []
for ch in expression:
if ch in OPEN:
st.append(OPEN[OPEN.index(ch)])
if ch in CLOSE:
if len(st) == 0:
return False
if st.pop() != OPEN[CLOSE.index(ch)]:
return False
return len(st) == 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"
Nov. 11, 2016
