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First solution in Clear category for Brackets by Angelika
def checkio(expression):
tab=[]
czy=True
for i in range(len(expression)):
if(expression[i]=='(' or expression[i]=='[' or expression[i]=='{'):
tab.append(expression[i]) #zrobić tu wrzucanie na stos
elif (expression[i]=='}' or expression[i]==']' or expression[i]==')'):
if len(tab)>0:
x=tab[len(tab)-1]
if ((expression[i]=='}' and x=='{') or (expression[i]==']' and x=='[') or (expression[i]==')' and x=='(')):
tab.pop()
else:
czy=False
break
else:
czy=False
break
if len(tab)>0:
czy=False
return czy
#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. 31, 2016