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First solution in Clear category for Brackets by bswitaj
def checkio(expression):
stack=[]
for i in range(len(expression)):
if expression[i]=='(' or expression[i]=='{' or expression[i]=='[':
stack.append(expression[i])
else:
if expression[i]==')':
if len(stack)>0:
x=stack.pop()
if x!='(':
stack.append(x)
return False
else:return False
if expression[i]==']':
if len(stack)>0:
x=stack.pop()
if x!='[':
stack.append(x)
return False
else:return False
if expression[i]=='}':
if len(stack)>0:
x=stack.pop()
if x!='{':
stack.append(x)
return False
else:return False
print(len(stack))
if len(stack)==0:
return True
if len(stack)>0:
return False
#return True or 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"
Oct. 14, 2017
