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First solution in Clear category for Brackets by erewhon
import re
def checkio(expression):
brackets = "".join(re.findall(r"[\(\)\[\]\{\}]", expression))
while True:
temp = brackets
brackets = brackets.replace("()", "")
brackets = brackets.replace("[]", "")
brackets = brackets.replace("{}", "")
if brackets == temp:
break
if len(re.findall(r"[\(\)\[\]\{\}]", brackets)) == 0:
return True
else:
return False
# 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"
Aug. 19, 2020
