Editorial for AAAA 1 P5 - Mirror Sort - Olympiads Online Judge

Editorial for AAAA 1 P5 - Mirror Sort

Remember to use this editorial only when stuck, and not to copy-paste code from it. Please be respectful to the problem author and editorialist.
Submitting an official solution before solving the problem yourself is a bannable offence.

Author: Sucram314

Observations

Notice that the absolute difference between any two adjacent elements in the array after any number of reflections will always be $1$ ~1~. Thus, this must be true in target array as well, otherwise the answer is automatically -1.

Now, observe that once two elements of the array become equal due to a reflection, they will always stay equal no matter what future reflections occur. This is because reflections act upon all elements in the array, and the behaviour of a specific value is independent of its index.

Thus, we should be careful to ensure that elements which become equal due to a reflection are also equal in the target array.

Additionally, if two elements become equal and are supposed to be equal in the target array, we only need to consider the behaviour of one of them throughout the rest of the construction, because if one of these elements reaches its target value, the other will be guarenteed to reach its equal target value, too.

Now, notice that when a reflection operation is performed on an increasing/decreasing sequence of consecutive non-negative integers (such as our starting array), the result is either another increasing/decreasing sequence of consecutive non-negative integers, or the absolute value "V-shape", where one side may be longer than the other.

If we apply the idea that we can throw away one of the elements in a pair of equal elements to the V-shape, we notice that all elements on the shorter side of the V-shape have a corresponding equal element on the longer side. Thus, assuming that these pairs of elements are supposed to be equal in the target array, we can now only consider solving the longer side of the V-shape, which is yet another increasing/decreasing sequence of consecutive non-negative integers.

In other words, there will always exist a "representative subarray" of our current array which contains the full range of values in our array from minimum to maximum in an increasing or decreasing sequence of consecutive non-negative integers. If this representative subarray is mapped to its corresponding subarray in the target array, the entire current array will be solved (assuming that elements outside of the representative subarray were paired correctly with elements inside the representative subarray). The length of the representative subarray is non-increasing under reflection operations since its left bound is non-decreasing and its right bound is non-increasing.

Construction

Let us begin our construction by focusing on creating the correct groups of equal elements, disregarding correct values.

Only reflections which create V-shapes can change the equality relationships between elements in the current array. They will allow us to pick some prefix/suffix of the current array and pair up elements to become equal in a palindromic fashion, reducing our problem to a subproblem represented by the longer side of the V-shape.

Thus, to reach the target array, we should search for odd length palindromes (with length at least 3) in its prefix/suffix, and slowly "fold" our way inwards until the remaining representative subarray corresponds to a increasing/decreasing sequence of consecutive non-negative integers in the target array, which can be solved trivially.

If there are no more palindromes in the prefix/suffix of the target array, but the corresponding subarray of the target array is not a strictly increasing/decreasing sequence of consecutive non-negative integers, then the target array is impossible to construct.

Proof that choosing an arbitrary prefix/suffix palindrome always works at any step

We only need to prove that the first reflection can correspond to an arbitrary prefix/suffix, and then we can apply the logic inductively to every following reflection to allow an arbitrary prefix/palindrome at every step.

Claim: If a construction exists, then a construction where the first reflection corresponds to an arbitrary prefix/suffix palindrome in the target array also exists.

Proof:

Let $x_1, x_2, \dots, x_M$ ~x_1, x_2, \dots, x_M~ be any sequence of reflections which is a valid construction. We know by definition this sequence of reflections maps $1,2,\dots,N$ ~1,2,\dots,N~ to $a_1,a_2,\dots,a_N$ ~a_1,a_2,\dots,a_N~.

If we change the first reflection to correspond to any prefix palindrome spanning from index $1$ ~1~ to $2k+1$ ~2k+1~, the problem reduces to mapping $0, 1, \dots, N-k-1$ ~0, 1, \dots, N-k-1~ to $a_{k+1}, a_{k+2}, \dots, a_N$ ~a_{k+1}, a_{k+2}, \dots, a_N~, since the elements to the left are now irrelevant.

We can apply a reflection about $N-k-1$ ~N-k-1~ to map $0, 1, \dots, N-k-1$ ~0, 1, \dots, N-k-1~ to $N-k-1, N-k-2, \dots, 0$ ~N-k-1, N-k-2, \dots, 0~. Then, we can apply a reflection about $k - (N-k-1)$ ~k - (N-k-1)~ to map $N-k-1, N-k-2, \dots, 0$ ~N-k-1, N-k-2, \dots, 0~ back to $k+1, k+2, \dots, N$ ~k+1, k+2, \dots, N~.

Now, we can simply apply our original sequence of reflections, $x_1, x_2, \dots, x_M$ ~x_1, x_2, \dots, x_M~ to map $k+1, k+2, \dots, N$ ~k+1, k+2, \dots, N~ to $a_{k+1}, a_{k+2}, \dots, a_N$ ~a_{k+1}, a_{k+2}, \dots, a_N~.

Thus, the target array will always be constructable after any arbitrary prefix palindrome is chosen for the first reflection.

We can show that this works similarly for suffix palindromes.

In conclusion, we do not need to worry about the target array starting constructible but becoming inconstructible after choosing an arbitrary prefix/suffix palindrome at any step.

Thank you dnialh_ for this fairly obvious proof that I could not find for some reason...

The number of reflections needed to reduce the problem down to a trivial increasing/decreasing sequence of consecutive non-negative integers in the target array is at most $N-2$ ~N-2~, since the representative subarray decreases in length by at least $1$ ~1~ after each reduction, and once the representative subarray becomes length $2$ ~2~, it is guarenteed to not contain any odd length palindromes (with length at least 3).

The trivial subproblem can be solved in at most $2$ ~2~ additional reflections.

In total, this construction takes at most $(N-2) + 2 = N$ ~(N-2) + 2 = N~ reflections, which fits exactly within the limit.

Subtask 1

This subtask was intended to reward $\mathcal{O}(N^2)$ ~\mathcal{O}(N^2)~ implementations of the construction described above.

This time complexity is most commonly achieved if the program searches for palindromes or directly simulates the reflection operation in $\mathcal{O}(N)$ ~\mathcal{O}(N)~ time. With at most $N$ ~N~ operations, this results in quadratic time complexity.

Time Complexity: $\mathcal{O}(N^2)$ ~\mathcal{O}(N^2)~

Subtask 2

In this subtask, a $\mathcal{O}(N)$ ~\mathcal{O}(N)~ solution is intended. The palindrome searching on the target array can be optimized by preprocessing for each index in the target array, the previous and next indices where the value at this index occurs (or if it does not occur). Combining this with a naive search from the center of the potential palindromes, the overall time complexity is linear, since the remaining representative subarray decreases in size proportionally to the amount of elements searched. Alternatively, you can use string hashing or Manacher's algorithm to check if subarrays are palindromic in $\mathcal{O}(1)$ ~\mathcal{O}(1)~ time.

For the reflection operation, we do not need to directly simulate it, as all the information we need during our construction is:

the left and right bounds of the remaining representative subarray,
the current value of the first element in the subarray, and
whether the subarray is increasing or decreasing,

all of which can be recalculated in constant time per reflection.

Time Complexity: $\mathcal{O}(N)$ ~\mathcal{O}(N)~

Bonus

Author Commentary

This problem was a complete accident. I was trying to make something related to Manhattan distance, and ended up with Ad Hoc final boss instead.

Also, if you understood anything from this editorial, try to write the checker for this problem! It's quite similar to the construction.

Note

Partial points are given to solutions which do not satisfy $M \le N$ ~M \le N~ but do satisfy $M \le 3N$ ~M \le 3N~. This was meant to reward solutions which repeatedly reset the representative subarray to an increasing sequence of non-negative integers for "easier" implementation (should take at most $2N$ ~2N~ operations), with some extra leniency.

However, this does not help very much in finding the construction in the first place, and if you do find the construction, you're likely to skip past this idea directly to the $M \le N$ ~M \le N~ solution anyway.

Honestly, I'm not sure how I would set up subtasks to hint towards the solution of this problem, even now.