correct me if I’m wrong, but O(n) would be the fastest since the comparisons(time) is directly proportional to word.length(). Unless I completely missed something.

so in order of time complexity(least to most complex): O(n) – O(n log n) – O(n^2)

correct me if I’m wrong, but O(n) would be the fastest since the comparisons(time) is directly proportional to word.length(). Unless I completely missed something.

so in order of time complexity(least to most complex): O(n) – O(n log n) – O(n^2)

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It is correct that O(*n*) is faster (grows more slowly forever) than O(*n* log *n*) and O(*n*^2).

I think the question was, however, what is the best order possible for an implementation of `IsIsogram()`

in the `BullCowGame`

program.

How can you determine that case (i.e., what is the best-possible big-O for this specific situation)?

Bonus: Can you think of a reason that there is an even better big-O in this specific case? Hint: can *n* grow indefinitely and if not, what might the big-O be?