Enable Javascript in your browser and then refresh this page, for a much enhanced experience.
First solution in Clear category for Brackets by hypehr96
def checkio(expression):
nawiasy = ""
out = True
for l in expression:
if l=='(' or l=='{' or l=='[':
nawiasy+=l;
if l==')' or l=='}' or l==']':
if(len(nawiasy)!=0):
znak=nawiasy[len(nawiasy)-1]
else:
out=False
break
if l==')' and znak=='(':
nawiasy = nawiasy[:-1]
elif l==']' and znak=='[':
nawiasy = nawiasy[:-1]
elif l=='}' and znak=='{':
nawiasy = nawiasy[:-1]
else :
out=False
break
if nawiasy!="": out=False
return out
#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"
Dec. 1, 2016
Comments:
