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First solution in Clear category for Brackets by werepire
def checkio(expression):
brackets=[]
for i in range(len(expression)):
b=expression[i]
if b in "([{":
brackets.append(b)
elif b in ")]}":
if len(brackets)==0:
return False
x=brackets.pop(len(brackets)-1)
if (b==')' and x=='(') or (b==']' and x=='[') or (b=='}' and x=='{'):
continue
else:
return False
if len(brackets)==0:
return True
return False
#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
