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First - Stack + Dict solution in Clear category for Brackets by Sarina
def checkio(expression):
stack = []
brackets = {')':'(', ']':'[', '}':'{'}
for c in expression:
if c in brackets.values():
stack.append(c)
elif c in brackets.keys():
try:
if stack.pop() != brackets[c]:
return False
except IndexError:
return False
return not 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"
assert checkio("(((1+(1+1))))]") == False
May 4, 2020
