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First solution in Clear category for Brackets by maciejmoscicki
def checkio(expression):
lista=[]
#print("*****")
for znaczki in expression:
if(znaczki=='(' or znaczki=='{' or znaczki=='['):
lista.append(znaczki)
#print(lista)
elif znaczki==')':
if(lista[len(lista)-1]=='('):
lista.pop()
else:
return False
elif znaczki=='}':
if(lista[len(lista)-1]=='{'):
lista.pop()
else:
return False
elif znaczki==']':
if(len(lista)==0):
return False
if(lista[len(lista)-1]=='['):
lista.pop()
else:
return False
if(len(lista)==0):
return True
else:
return False
#print(lista)
#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. 15, 2016
