Enable Javascript in your browser and then refresh this page, for a much enhanced experience.
recursion solution in Clear category for Brackets by EdinsonUwU
def entrada(expression):
brackets = ['(', ')', '{', '}', '[', ']']
listaBrackets = []
for i in expression:
if i in brackets:
listaBrackets.append(i)
return listaBrackets
def checkio(lista):
lista = entrada(lista)
dictionary = {'(' : ')', '[' : ']', '{' : '}'}
if lista == []:
return True
if (len(lista)%2) != 0:
return False
try:
for i in range(0,len(lista)):
if lista[i] in dictionary:
if lista[i+1] == dictionary[lista[i]]:
lista.pop(i)
lista.pop(i)
return checkio(lista)
except:
return False
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"
Feb. 10, 2021
