problem with None

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I’m doing an exercise here and the desired output would be Yes in this case but it’s returning me None. Could anyone help me? vlw

def alphabet (s,t,k):


    for i in range(0,len(s)):
        if s[i] != t[i] or s == 0: #o i indica o numero de chars que o s e o t tem em comum
            prefix = i
            break
        
            e = len(s) - prefix - len(t) - prefix

            if len(s) + len (t) < k:
                return 'Yes'
            elif e <= k and (e % 2) == (k % 2):
                return 'Yes'
            else:
                return 'No'






print (alphabet ('qwerasdf', 'qwerbsdf', 6))
python

  • Every function that does not have a Return explicit, automatically assumes return None. In your case, not going into the first if of the function, the return will be None

1 answer

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The problem occurs because the function has no return to the print, when the code hits the break, the solution would be to add a return in the function, for example:

def alphabet (s,t,k):

    for i in range(0,len(s)):
        if s[i] != t[i] or s == 0: #o i indica o numero de chars que o s e o t tem em comum
            prefix = i
            break

            e = len(s) - prefix - len(t) - prefix
    
            if len(s) + len (t) < k:
                return 'Yes'
            elif e <= k and (e % 2) == (k % 2):
                return 'Yes'
            else:
                return 'No'

    return 'No | Yes'

print (alphabet ('qwerasdf', 'qwerbsdf', 6))

In that case you passed, would return 'No | Yes'.

Obs: Another detail is in relation to s == 0 here you are just comparing the content of the list with zero, this is exactly what you want to do?

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