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A is an array containing at most 105 integers.

We have to do 2 kinds of operations on this array in log(N) complexity (where, N= number of elements in A).

Operation 1, given v,i,j we have to add v to A[k] (i<=k<=j).

Operation 2, given i & j calculate ( A[i] * A[i+1] * A[i+2] * .... * A[j] ) % M. (M is a prime, and will be same for all operations).

There will be almost 105 operations to be made.

If it's not possible in log(N), then what is the best possible complexity to do the operations?

Bidhan Roy
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    For operation 1, do you really mean `add v` or `multiply by v`? For `multiply by v`, my solution can be adapted more easily. I still think segment trees are a viable option for a `O(log N)` solution, but I don't see how to do Operation 1 so far. You tagged this with spoj, can you link to the spoj problem? – IVlad Aug 21 '13 at 15:37
  • @IVlad I also figured out that for **multiply by v** it can be done by segment tree or binary indexed tree. This is not a specific spoj problem, but knowing the trick will help me in solving several problems from spoj. – Bidhan Roy Aug 21 '13 at 16:17
  • Trying to work this one out; I am also convinced that sublinear is possible. I haven't fully worked out the issues yet, but I'm willing to bet that it'll involve the fact that (integers from `0` to `M-1`, addition modulo `M`, multiplication modulo `M`) forms a field. – Dennis Meng Aug 21 '13 at 16:51

2 Answers2

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Since it looks like you have to access all the elements in the range [i, j], the complexity depends on the linear size of that range,

Paul Evans
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  • Well, the first operation can be done in log(K) complexity by using some data-structures, where K is the linear size of that range. May be the second one can also be done in the same complexity using data-structure techniques. – Bidhan Roy Aug 21 '13 at 14:32
  • Then find a way to do this without having to access all the elements. :P – Dennis Meng Aug 21 '13 at 16:53
  • @DennisMeng - while I will definitely try to do that, me being unable to do so does not prove that it is not possible. – IVlad Aug 21 '13 at 22:03
  • @IVlad I was actually talking to Paul, but I also want to make a distinction between having to access each individual element on its own (which *would* cause linear-time worst-case), and being able to use some processing to avoid having to do so. – Dennis Meng Aug 21 '13 at 22:07
  • Yes, you can cache all the ranges in `O(n^2)` (with n ~ 10^5) complexity and then run the above in `O(1)`. I really don't see this gets *around* the fact that it's basically an `O(n)` problem. – Paul Evans Aug 22 '13 at 00:44
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It's possible that j-i is on the order of N, and you've got to change each of them. That makes any algorithm faster than O(N) impossible, as Paul said. K is not a parameter of the problem, it's just a variable, therefore log(K) in Bidhan's answer makes no sense.

Now, if the question would be not about time complexity proper, but about the height of the tree of massively parallel operations (such as you have on CUDA, for example), then, given enough threads one would trivially perform Operation 1 in O(1) due to independence of all operations, and Operation 2 in O(log(N)) by multiplying mod M pairs of adjacent elements (requires 100000/2 threads), then pairs of adjacent results, etc, until the answer is reached.

This was, however, not the question. Barring massively parallel computing the complexity of each operation is O(N), no matter how you perform that.

Michael
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  • "you've got to change each of them" does not follow. Even if faster than `O(N)` were impossible (which I highly doubt), this is not a proof. You don't necessarily have to change each of them, it might be sufficient to store certain information regarding the fact that they have been changed and still be able to answer the Operation 2 queries. – IVlad Aug 21 '13 at 15:42
  • Yes, as **IVIad** said, you don't need to access every element to calculate the answer. And, the log(K) makes perfect sense to someone with some knowledge of segment tree. – Bidhan Roy Aug 21 '13 at 16:20