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Second (revised solution) solution in Clear category for Brackets by vlad.n
def checkio(data):
brack_dict = {")": "(", "]": "[", "}": "{"}
stack = []
for d in data:
# found opening bracket
if d in brack_dict.values():
stack.append(d)
# found closing bracket
elif d in brack_dict:
if stack and stack[-1] == brack_dict[d]:
stack.pop()
else:
return False
return stack == []
#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"
Jan. 11, 2018
