Enable Javascript in your browser and then refresh this page, for a much enhanced experience.
"Brackets" solution in Clear category for Brackets by iskenderunbtr
def checkio(expression):
phrase="".join([x for x in expression if x in "(){}[]"])
count=0
maxx=len(phrase)//2
while True:
phrase=phrase.replace("()", "").replace("{}", "").replace("[]", "")
count+=1
if count>maxx:break
return not phrase
#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"
June 13, 2020
Comments: