Enable Javascript in your browser and then refresh this page, for a much enhanced experience.
First solution in Clear category for Brackets by Ephemeral
def checkio(expression):
open_b = '({['
close_b = ')}]'
l = []
for c in expression:
if c in open_b:
l.append(c)
elif c in close_b:
if len(l)==0 or open_b.index(l.pop()) != close_b.index(c):
return False
return l == []
#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"
March 16, 2019
