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First solution in Clear category for Brackets by Lukeram
list = "{(["
lol = "})]"
def check (i, n, a, b):
if i == lol[n]:
if not a[-1] == list[n]:
return False
elif not b == False :
return True
def checkio(expression):
a=""
b = True
for i in expression:
if b == False:
break
if i in list:
a=a+i
elif i in lol:
if len(a) < 1:
b=False
else:
b=check(i, lol.find(i), a, b)
a=a[:len(a)-1]
c=(len(a) == 0)
return b and c
#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"
Nov. 20, 2016
