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Straight solution in Clear category for Brackets by YuriyM
# lists of corresponding open and closed brackets.
# order has matter
OPENS = ['(', '{', '[']
CLOSES = [')', '}', ']']
def checkio(expression):
expected_closed = [] # collect expected closed bracket for corresponding open one
for l in expression:
if l in OPENS:
expected_closed.append(CLOSES[OPENS.index(l)])
elif l in CLOSES:
# verify if closed bracket equal to the expected one
if expected_closed and expected_closed[-1] == l:
expected_closed.pop(-1) # remove the verified one
else:
return False
# Check if all open brackets are closed
if expected_closed:
return False
return True
#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"
May 27, 2020
