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Brackets solution in Clear category for Brackets by mz97
def checkio(expression):
e=expression
l=""
n=-1
for i in range(len(e)):
if (e[i]=='(') or (e[i]=='[') or (e[i]=='{'):
l=l+e[i]
n+=1
elif e[i]==')':
if n==-1 or l[n]!='(':
return False
n-=1
l=l[:-1]
elif e[i]==']':
if n==-1 or l[n]!='[':
return False
n-=1
l=l[:-1]
elif e[i]=='}':
if n==-1 or l[n]!='{':
return False
n-=1
l=l[:-1]
if n==-1:
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"
Oct. 13, 2016