I would argue that this is still brute-force. Since every array element (at least till a mismatch is found) must still be examined. In this case "brute-force" = O(n) (time is proportional to the size of the list). The technique presented above has the performance characteristic that its speed is directly proportional to the list size so IMHO it is still brute-force (or at least performance-equivalent to brute-force).

However, if you were to do something like populate your list elements into a hash as keys at the same time as you added them to a list then determining that they were all the same would be O(1), better than brute-force. However, this would slow-down your "add an element to the list" function.

In reply to RE: RE: Re: All array elements the same? by lhoward
in thread All array elements the same? by Grumpy

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