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They say "if it works it aint stupid". But this seems pretty stupid. solution in Creative category for Brackets by Dominik_Witczak
def checkio(expression):
sb1=0 #(
sb2=0 #[
sb3=0 #{
eb1=0 #)
eb2=0 #]
eb3=0 #}
last = []
for i in expression:
if i == "(":
sb1 += 1
last.append(1)
if i == "[":
sb2 += 1
last.append(2)
if i == "{":
sb3 += 1
last.append(3)
if i == ")":
eb1 += 1
try:
if last.pop() != 1:
return False
except:
return False
if i == "]":
eb2 += 1
try:
if last.pop() != 2:
return False
except:
return False
if i == "}":
eb3 += 1
try:
if last.pop() != 3:
return False
except:
return False
if sb1 != eb1 or sb2 != eb2 or sb3 != eb3:
return False
return 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"
Oct. 27, 2017
