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Brackets solution in Clear category for Brackets by Adrian_W
def checkio(expression):
stack=[]
for i in expression:
if i in "{[(":
stack.append(i)
elif i ==')':
try:
if stack.pop()!='(':
return False
except:
return False
elif i ==']':
try:
if stack.pop()!='[':
return False
except:
return False
elif i =='}':
try:
if stack.pop()!='{':
return False
except:
return False
if len(stack)>0:
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"
Oct. 11, 2016
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