Enable Javascript in your browser and then refresh this page, for a much enhanced experience.
First solution in Uncategorized category for Brackets by mkobylka
def checkio(expression):
BRACKETS="()[]{}"
openBrackets=[]
for b in expression:
if BRACKETS.count(b)==0: continue # No bracket character
if "([{".count(b): openBrackets+=b # Adding open bracket
else: # if char is a close bracket...
if not openBrackets: return False # Empty open brackets array case
lastBracketIndex=BRACKETS.index(openBrackets[-1])
if b!=BRACKETS[lastBracketIndex+1]: return False # Different type of bracket case
else: openBrackets=openBrackets[:-1] # Removing bracket from array...
if openBrackets: return False # Some open bracket remaining case
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"
Dec. 18, 2015
