Is this list comparison in Python unnecessary?

Question

The Count and Compare Anagram solution provided at interactivepython.org checks iterates through the lists for a final time checking if the count for each ASCII value is the same.

j = 0
stillOK = True
while j<26 and stillOK:
    if c1[j]==c2[j]:
        j = j + 1
    else:
        stillOK = False

return stillOK

Why not use the comparison operator?

return (c1 == c2)

Full code:

def anagramSolution4(s1,s2):
    c1 = [0]*26
    c2 = [0]*26

    for i in range(len(s1)):
        pos = ord(s1[i])-ord('a')
        c1[pos] = c1[pos] + 1

    for i in range(len(s2)):
        pos = ord(s2[i])-ord('a')
        c2[pos] = c2[pos] + 1

    j = 0
    stillOK = True
    while j<26 and stillOK:
        if c1[j]==c2[j]:
            j = j + 1
        else:
            stillOK = False

    return stillOK

print(anagramSolution4('apple','pleap'))

Edited to add:

I've tested with:

anagramSolution4('abc','cba') #returns True
anagramSolution4('abc','cbd') #returns False
anagramSolution4('abc','cbah') #returns False

..they all pass. What would be an appropriate test to show c1==c2 fails?

Answer 1

Using == on the two lists would produce the same result, but would also hide some implementation details. Given that the script comes from a website used for learning, I guess this is for learning purposes.

Also, I see that in the webpage you are asked some questions about complexities. Well, using c1 == c2 instead of the loop would probably mislead some people and make them think that the operation is O(1) instead of O(min(len(c1), len(c2)))^[1].

Finally, note that there are many languages which have no notion of lists.

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^[1] This isn't necessarily true either, as the two lists may contain items with custom and complex __eq__() methods.