Enable Javascript in your browser and then refresh this page, for a much enhanced experience.
First solution in Clear category for Brackets by bartorbard
def textOut (text):
brackets = "(){}[]"
for eachChar in text:
if eachChar not in brackets:
text = text.replace(eachChar, "")
return text
def function2 (text):
text = str(text)
text = textOut (text)
textRep = text
idx2=0
while (idx2 < len(text)/2):
idx = 0
brack = [ "()", "[]", "{}"]
while (idx<3):
textRep = textRep.replace(brack[idx],"")
idx = idx+1
idx2 = idx2+1
return textRep
def checkio(expression):
text = expression
textResult = function2(text)
if len(textResult) == 0:
textResult = True
else:
textResult = False
return (textResult)
#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. 27, 2016
