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Stosz solution in Uncategorized category for Brackets by kamilk
def checkio(expression):
stosz = []
for c in expression:
if c=='(': stosz.append(')')
if c=='[': stosz.append(']')
if c=='{': stosz.append('}')
if c==')' or c=='}' or c==']':
if len(stosz)==0:
return False
if stosz[-1]!=c:
return False
stosz.pop()
if len(stosz)!=0:
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"
Dec. 18, 2015
