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First solution in Clear category for Brackets by bocian.michal
def checkio(expression):
stack = [""]
expr = True
count = 0
for c in expression:
if c in "([{":
stack.append(c)
count+=1
elif c == ")":
count-=1
if stack.pop() != "(":
expr = False
break
elif c == "]":
count-=1
if stack.pop() != "[":
expr = False
break
elif c == "}":
count-=1
if stack.pop() != "{":
expr = False
break
if count != 0:
expr = False
return expr
#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. 6, 2016
