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Stack solution in Clear category for Brackets by Ambulanss
# migrated from python 2.7
brac1 = ( "{" , "[" , "(")
brac2 = ( "}" , "]" , ")" )
def checkio(expression):
stack = []
for i in expression:
if i in brac1:
stack.append(brac1[brac1.index(i)])
if i in brac2:
if not len(stack):
return False
if stack.pop() != brac1[brac2.index(i)]:
return False
return not len(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"
Nov. 20, 2016
