Enable Javascript in your browser and then refresh this page, for a much enhanced experience.
First solution in Clear category for Brackets by KrzysztofP
def checkio(expr):
opening=set('([{')
close=set('}])')
match=set([ ('(',')'), ('[',']'), ('{','}') ])
stack=[]
for char in expr:
if char in opening:
stack.append(char)
elif char in close:
if len(stack)==0:
return False
lastOpen=stack.pop()
if (lastOpen, char) not in match:
return False
return len(stack)==0
#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"
Nov. 1, 2016
