### bicsi's blog

By bicsi, history, 20 months ago,

... without doing much of anything.

I've been obsessed for the last two days about problem Data Centers from EGOI 2022. In particular, it has a subtask where $1 \leq N \leq 10^5$ and $1 \leq S \leq 5000$, where it seems $O(NS)$ wouldn't fit TL. I've been conjecturing that using treaps would amortize to good complexity, and after two days of burning grey matter, I've finally found a proof, and it's pretty cool to make a blog out of it.

The problem basically starts with an array $[v_1, v_2, ..., v_N]$, where $1 \leq v_i \leq V$ for all $i$. It then asks us to iteratively simulate the operation: "Subtact $x_i$ from the biggest $m_i$ elements.". All this while keeping the array sorted.

We'll show that just using treaps (or any BBST for that matter) yields a good amortized complexity. More exactly, we'll prove $O((N + Q) \log V \log N)$, and it should actually be better in practice.

### Naive solution with treaps

Of course, the "naive" solution (which is much of what we'll implement here), does three things:

1. Extract the $m_i$ suffix of the sorted array (we'll call it $B$, and the remainder $A$)
2. Subtract $x_i$ from all elements of $B$
3. Merge the resulting sequences $A$ and $B$

Using treaps, it's easy to see that steps 1 and 2 can be done in $O(\log N)$ via split and lazy propagation. Step 3 is the heaviest, requiring potentially $O(N)$ and maybe even $O(N \log N)$ steps depending on implementation.

#### Interweaving number

For a given merging process of sequences $A$ and $B$, let $k$ be the number of interweaving segments. See example below:

A = {2, 5,          15,   25, 30}
B = {      7, 9, 12,    21      }


In the example above, we have $k = 5$ interweaving segments (of lengths $2$, $3$, $1$, $1$, and $2$ respectively).

First, it is rather straightforward to write an implementation of merge on BBST that has complexity $O(k \log N)$ (via $k$ binary searches and $k$ splits).

Secondly, there can be at most $O((N + Q) \log V)$ merging interweaving segments over an array of size $N$ and $Q$ queries.

### Potential method

For a given (increasing) sequence $A = [x_1, x_2, ..., x_n]$, let's define:

$\pi(A) = \sum_{i=1}^{n-1} \log(x_{i+1} - x_{i})$

(all logarithms are base-2)

Let's suppose we are merging two sequences $A$ and $B$, and there are $k$ interweaving segments. See example below:

             <-d1->         <--d2-->      <--d3-->       <---d4--->
A = {[--a1--],                      [-a2-],                        [--a3--]}
B = {              [---b1---],                     [-b2-]                  }


In the picture above, the segments [-----] are placeholder to some sequence of elements. Numbers $d_i$ are the distances between numbers, e.g. $d_1 = first(b_1) - last(a_1)$.

The difference between potentials before and after merge $\Delta \pi = \pi (A) + \pi (B) - \pi (A \cup B)$ is:

$\Delta \pi = (\log (d_1 + b_1 + d_2) + \log (d_2 + a_2 + d_3) + \cdots + \log (d_{k-1} + a_{k/2} + d_{k})) - (\log d_1 + \log d_2 + \cdots + \log d_k) \quad (1)$

Nonetheless, we can see that:

$\Delta \pi \geq (\log (d_1 + d_2) + \cdots + \log (d_{k-1} + d_k)) - (\log d_1 + \cdots + \log d_k) \quad (2)$

(more naturally, the "worst case" is when doing a Faro shuffle, i.e. interweaving element by element).

However, because $\log$ is concave, we can use that $\log (\frac{a + b}{2}) \geq \frac{\log a + \log b}{2}$. Reformulating, we obtain:

$\log(a+b) \geq 1 + \frac{1}{2}(\log a + \log b) \quad (3)$

We'll add and subtract $\log (d_k + d_1)$ to inequality $(2)$, use inequality $(3)$, and simplify everything to finally obtain:

$\Delta \pi \geq k - \log (d_1 + d_k)$

(thanks freak93 for the easy one-liner about concavity of $\log$ )

The initial potential of the sequence is bounded by $N \log V$, and each merge operation adds at most $\log V$ to the potential, yielding a total of $(N + Q) \log V$ potential. Then, the number of interweavings is at most $(N + Q) \log V$.

### Extra

It's not hard to see the power of this result. In essence, you can do all sorts of operations like: adding/subtracting some value $x$ to an arbitrary subsequence, dividing subsequence by some value, square rooting subsequence, and so on. All while keeping the sequence sorted.

Also, doing $\log N$ binary searches on treaps might be inconvenient. Another approach that is easier to code and faster is to use the treap union function (source: Wikipedia):

function union(t1, t2):
if t1 = nil:
return t2
if t2 = nil:
return t1
if priority(t1) < priority(t2):
swap t1 and t2
t<, t> ← split t2 on key(t1)
return join(union(left(t1), t<), key(t1),
union(right(t1), t>))


### Conclusions

• It feel like this is a bit more general method to this
• Replace join treap operation in your library with merge
• I feel like this trick is well known in some countries...
• +150

 » 20 months ago, # | ← Rev. 2 →   +10 There is a problem on CF that I believe can be solved with the same trick: 702F - T-Shirts
 » 20 months ago, # | ← Rev. 2 →   +26 This proof can be found here: String matching in Lempel-Ziv compressed strings.Also, this paper gives a faster, $\mathcal O(\log V)$ amortized time algorithm: Mergeable Dictionaries With Shifts.
 » 6 months ago, # |   +33 Why does the given implementation work in $O(k \log n)$? It seems that when $\max{t_1} < \min{t_2}$, it is $O(\log^2 n)$ instead of $O(\log n)$.
•  » » 4 weeks ago, # ^ |   0 Yes, you are right. Each split is $O(\log n)$ and split is called $O(\log n)$ times. To avoid this and optimize the time complexity to $O(\log n)$, we can store the max and min keys for each subtree. Judging by these max and min keys, when we are splitting, if the entire tree is to the left or to the right of the split key, we just quit splitting to save time.
 » 6 months ago, # |   0 911G — Mass Change Queries Also can be solved with this
 » 6 months ago, # |   -10 Yeah in a chinese data structure called FHQ-treap, spliting and merging on treaps are basic operations.