By time limit exceeded I assume you mean they wanted to the running time of the algorithm to be faster? If so I think you have a good starting point in your second solution.
I don't have the answer, but I would probably think about why that algorithm is so slow, which might depend on the size of the numbers being multiplied or the size of the arrays themselves.
If the numbers are large maybe there is a way to know something about the difference of the two products without computing the actual products.
For example, without calculating anything I know that the product of [13,27] is less than [14,28], and both will be closer to each other than [13,28,92].
The other thing that comes to mind is you basically have have (N-2) possible values of X, but maybe you don't need to examine all of them? For these kinds of problems I find it helpful to work through the algorithm manually and see where I could cut corners.
My thought process goes something like this:
If you start with N=1 then you have only 1 number in the left array, and all the other numbers in the right array.
This makes me wonder if N=1 is always the worst case, but there is a counterexample: the number on the left hand side could be 9 and all the other numbers could be very low, like [3,1,1,1,1,3].
Maybe the array could contain zeros as well
If there are multiple zeros you can choose any X between 2 of the zeros and then the difference will be zero
If there is a single zero, one of the products will be zero and the other will be non zero. Then you can choose X to make the non zero part is as small as possible
assuming the numbers are integers then the smallest non zero part will be a single number: either X=1 or X=N-2
So in special cases you can avoid multiplication entirely
What about a non-special case: [3,2,7,1,5]?
Let's say we do start with N=1: I've multiplied all the numbers together and I divide by 3. Did I need to know the full product for that?
I've done 3x2x7x1x5/3 so I needed most of it but the two 3s cancelled out, so I didn't need the full product
And for N=2 I've done 3x2x7x1x5/3/2. This doesn't seem like the easiest way to calculate that
If I have N>1 then I have minDiff target I need to beat - in this case I case I don't need to know the exact diff, if I can prove it's bigger than minDiff
Given two arrays [A_1,A_2...A_X], [A_X+1 ... A_N] can I break the calculation into parts and do that check part way through?
If I calculate the product of one array as Z then the product of the second array must be less than
minDiff+Z to be worth calculating
If the numbers are integers, and we've already ruled out zeros, then calculating the product step by step will cause the number to get bigger each iteration
So we can stop early if we exceed minDiff
Also, if we get to this point and the second array is smaller than the first, then increasing X won't help, so we can terminate early and return minDiff
I doubt this thought process would get me anywhere in an interview setting, but I think it would give me some ideas for different approaches. Even if these ideas don't work, I think I'd learn something about the problem from exploring them.
Also now I'm wondering whether my choice of programming language was the reason for timeout.
Usually in most online platforms, the timeout would be different for different programming languages right?
But the time and memory constraints for this problem was fixed at something like 2s and 350 MB. (I'm unable to check now. The challenge has ended)
And the input was quite large. When I used the / operator, I got an error saying can't divide such a huge float or something like that. I had to use // instead to do integer division.
So maybe, I should have tried in C++/Go too.
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By time limit exceeded I assume you mean they wanted to the running time of the algorithm to be faster? If so I think you have a good starting point in your second solution.
I don't have the answer, but I would probably think about why that algorithm is so slow, which might depend on the size of the numbers being multiplied or the size of the arrays themselves.
If the numbers are large maybe there is a way to know something about the difference of the two products without computing the actual products.
For example, without calculating anything I know that the product of [13,27] is less than [14,28], and both will be closer to each other than [13,28,92].
The other thing that comes to mind is you basically have have (N-2) possible values of X, but maybe you don't need to examine all of them? For these kinds of problems I find it helpful to work through the algorithm manually and see where I could cut corners.
My thought process goes something like this:
minDifftarget I need to beat - in this case I case I don't need to know the exact diff, if I can prove it's bigger thanminDiffminDiff+Zto be worth calculatingminDiffminDiffI doubt this thought process would get me anywhere in an interview setting, but I think it would give me some ideas for different approaches. Even if these ideas don't work, I think I'd learn something about the problem from exploring them.
That's some good insight into the problem!
Also now I'm wondering whether my choice of programming language was the reason for timeout.
Usually in most online platforms, the timeout would be different for different programming languages right?
But the time and memory constraints for this problem was fixed at something like 2s and 350 MB. (I'm unable to check now. The challenge has ended)
And the input was quite large. When I used the / operator, I got an error saying can't divide such a huge float or something like that. I had to use // instead to do integer division.
So maybe, I should have tried in C++/Go too.