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First solution in Clear category for Brackets by sts10131
def checkio(expression):
brackets=""
for x in expression:
if x=='(':
brackets+='('
elif x=='[':
brackets+='['
elif x=='{':
brackets+='{'
elif (x==')' or x==']' or x=='}') and brackets=="":
return False
elif x==')':
if brackets[-1]=='(':
brackets=brackets[0:len(brackets)-1]
else:
return False
elif x==']':
if brackets[-1]=='[':
brackets=brackets[0:len(brackets)-1]
else:
return False
elif x=='}':
if brackets[-1]=='{':
brackets=brackets[0:len(brackets)-1]
else:
return False
if brackets=="":
return True
return False
#These "asserts" using only for self-checking and not necessary for auto-testing
if __name__ == '__main__':
print(checkio("(((1+(1+1))))]"))
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"
Dec. 27, 2016