It seems intuitive that there's no list data structure which has $\mathcal{O}(1)$ worst case time complexity for random access and a worst case complexity better than $\mathcal{O}(n)$ for insertion: if insertion is allowed to affect only a small part of the list, then there's no way for it to always keep the entire list in a structure that allows for constant time element access.

Is there any proof (or counterexample) of this?

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    $\begingroup$ When you say "insertion", what do you mean? Insertion at a specific index? $\endgroup$ – orlp Dec 5 '18 at 6:39
  • $\begingroup$ @orlp Yes, insertion at a specific index. $\endgroup$ – Stefan Dec 5 '18 at 6:40
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    $\begingroup$ In which computation model? This is important, because in real world if you keep connecting more and more memory to your computer (as big-Oh assumes), you can't expect constant-time random access to the memory. $\endgroup$ – Dmitri Urbanowicz Dec 5 '18 at 8:41
  • $\begingroup$ @DmitriUrbanowicz If you don't allow constant-time random access, you're just never gonna get O(1) here, so that's a bit ridiculous. Of course OP assumes random access. $\endgroup$ – Pål GD Jan 5 at 9:19
  • $\begingroup$ Do you have to have O(1) worst case, or is O(1) amortized acceptable? $\endgroup$ – Pål GD Jan 5 at 9:20

Yes, you can have both as fast as possible random access and faster-than-full-rebuild insertion time.

I assume you know how dynamically growing arrays work. Also, I assume you know how to make them work in $O(1)$ worst-case. These techniques allow us to focus on the problem for lists of limited size only.

Let $n$ be the maximal size (capacity) of the list. Let’s divide the list into $O(\sqrt{n})$ blocks of equal size $B$ (which is $O(\sqrt{n})$ too). Let also $b_j$ denote a cyclic shift of the $j$-th block.

When you need to access $i$-th element of the list, you calculate its position as follows in $O(1)$:

$$f(i) = (b_j + i \mod B) + B \times j$$

where $j = \lfloor \frac{i}{B} \rfloor$.

When you need to insert something into $i$-th position, you first make a cyclic shift of everything from $j$-th block to the end:

  1. Decrement all $b$ values starting from $j$.
  2. Swap the elements on the boundaries of consequtive blocks, so that every element is in its right place.
  3. Then you fix $j$-th block by rebuilding it from scratch.

This gives you $O(\sqrt{n})$ insertion complexity.


As accessing to an element is in $O(1)$, it would be something like an array. And, if this assumption is true, your discussion would be near the concept of "Dynamic Array". It's insertion to the array is $O(1)$ and amortized cost of $n$ insertion to the data structure is $O(n)$. It might be helpful for what you want to reach.

  • $\begingroup$ I didn't downvote you, and don't know who did, but inserting into the beginning or to the middle in a dynamic array is O(n). Inserting at end is amortized O(1). I think OP wanted arbitrary insert to be (amortized?) O(1). $\endgroup$ – Pål GD Jan 5 at 9:17

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