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First solution in Clear category for Brackets by krispolska
# migrated from python 2.7
def checkio(e):
openBrackets = "[({"
closeBrackets = "])}"
opened = []
for l in e:
if l in openBrackets:
opened.append(l)
elif l in closeBrackets:
if opened:
if openBrackets.index(opened.pop()) == closeBrackets.index(l):
continue
return False
return False if opened else True
#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"
May 16, 2016
