Not sure I follow, to be honest. In the problem, we need to calculate, for example, the product of 1*2*3*4 two separate times, meaning that that particular result is required exactly twice. The second time, we don't have to recalculate it because we've already cached it. Each request is not unique—it happens twice.
It's true that we only read from the cache once per item, but does it make a difference whether we're reading each element from the cache once or 500 times?
Well, yes, I think it kind of does, because it governs what's an appropriate technique. Certainly in the case of the forward direction, all you've actually implemented is an accumulator, and in a particularly inefficient way as far as memory is concerned. And perhaps I got a bit muddled because you only presented the forward direction - the backwards direction probably does require memoisation (Edit: No, it doesn't).
The real point to this interview question, I believe, is that the most efficient solution requires different approaches for the forward and backward directions.
OK, I've had a proper chance to think about this (when I'm not posting from work/on the move), and Memoization is definitely not an ideal optimisation technique here. The problem is that what you really want is to not have to recompute things all the time, and so you want to have partial results. However, because of the form of these partial results, there's no requirement to actually store them in either a precomputed array or by memoization. It's trivial to use a pair of accumulators, as in the following code (Python, because that's what I normally use)
output_array =  * len(input_array)
reverse_acc = 1
for index, val in enumerate(reversed(input_array)):
output_array[len(input_array) - index - 1] = reverse_acc
reverse_acc *= val
forward_acc = 1
for index, val in enumerate(input_array):
output_array[index] *= forward_acc
forward_acc *= val
This version is likely to be as fast as you can get the algorithm, is pretty simple, and uses the bare minimum of additional memory (for clarity the two accummulators are seperate; they can be combined, meaning that the overhead needed for this algorithm is a single integer + the output array).
If you don't object, I might write this up into an object lesson on how to tell if the optimizations people choose to apply are actually appropriate optimizations. As I was saying, if you're only reading your memoized value once, it's unlikely to be the correct technique.
Thanks for all your insights, David. This is a really awesome solution. It didn't quite occur to me that I could have two separate accumulator loops (forward and reverse), and then combine the results like this.
You're definitely welcome to write this up in a post. That would be super interesting.
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