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First solution in Clear category for Brackets by JohnOZ
def checkio(str):
clean = [i for i in str if i in '[](){}']
clean = ''.join(clean)
while '()' in clean or '{}' in clean or '[]' in clean:
clean = clean.replace('[]','').replace('{}','').replace('()','')
if len(clean) == 0:
return True
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"
April 30, 2015
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