Editorial : Make it Palindrome 2

The solution has three short ideas stacked on top of each other:

Reduce to pair-gaps. Only the gaps between symmetric pairs matter.
Move to the difference array. Range-add problems usually become point-change problems after differencing.
Minimum-cost signed representation. A clean bookkeeping finishes the problem in one sorted pass.

We explain each step with the first sample in hand:

$$ N = 4,\; M = 5,\; A = [0,\, 3,\, 1,\, 2]. $$

1. Only the pair-gaps matter

A palindrome means $A_i = A_{N+1-i}$ for every $i$. The middle element (when $N$ is odd) pairs with itself, so it is automatically fine and we ignore it.

For each of the $h = \lfloor N/2 \rfloor$ pairs, define the cyclic gap

$$ b_i \;=\; (A_i - A_{N+1-i}) \bmod M \;\in\; \{0,1,\ldots,M-1\}. $$

The palindrome condition becomes: every $b_i = 0$.

In the sample, pairs are $(0, 2)$ and $(3, 1)$, giving $b = [3,\; 2]$.

2. Operations in $A$ become range-add operations on $b$

Take any operation on $A$: “add $1$ mod $M$ to every index in $[l, r]$.” For a pair $(i,\; N+1-i)$, exactly one of four things happens:

Range covers both elements → both go up by $1$ → $b_i$ unchanged.
Range covers only $A_i$ → $b_i$ goes up by $1$ (mod $M$).
Range covers only $A_{N+1-i}$ → $b_i$ goes down by $1$ (mod $M$).
Range covers neither → $b_i$ unchanged.

Because symmetric pair-indices move in opposite directions as $i$ grows, the set of pairs that change turns out to be a contiguous interval of pair-indices, and all pairs in that interval change the same way (all $+1$ or all $-1$):

An operation entirely in the left half adds $+1$ to some contiguous range of $b$.
An operation entirely in the right half adds $-1$ to some contiguous range of $b$.
A middle-crossing operation adds $+1$ or $-1$ to a contiguous range of $b$ (or changes nothing).

Conversely, every “$+1$ to $b$ on range $[p, q]$” can be realized as operation $[p, q]$ on $A$ (stays in the left half), and every “$-1$ to $b$ on range $[p, q]$” can be realized as the mirrored operation $[N+1-q,\; N+1-p]$ on $A$ (stays in the right half).

So the problem is exactly equivalent to:

Starting from a vector $b \in (\mathbb{Z}/M)^h$, one operation adds $+1$ or $-1$ (mod $M$) to a contiguous range of $b$. Minimum operations to reach the zero vector.

3. Difference array: range-add becomes unit-transfer

Classic trick. Pad $b$ with zeros on both ends, $(0, b_1, \ldots, b_h, 0)$, and take consecutive mod-$M$ differences:

$$ d_0 = b_1,\qquad d_i = (b_{i+1} - b_i) \bmod M,\qquad d_h = (0 - b_h) \bmod M. $$

So $d$ has $m = h + 1$ entries in $[0, M)$.

One range-add on $b$ only changes $d$ at the two ends of the range — it takes one unit from one position of $d$ and gives it to another. So our reformulated problem becomes:

Use unit-transfers between positions of $d$ to make every entry equal to $0$ mod $M$. Minimize the number of transfers.

In the sample, $b = [3, 2]$ extends to $(0, 3, 2, 0)$, so

$$ d = \bigl((3-0)\bmod 5,\; (2-3)\bmod 5,\; (0-2)\bmod 5\bigr) = [3,\; 4,\; 3]. $$

4. The integer sum of $d$ is a multiple of $M$

If we wrote each difference without wrapping, the “true” integer differences would telescope to $0$ (because $b$ is padded with $0$ on both sides). Wrapping each into $[0, M)$ adds some nonnegative multiple of $M$. Therefore

$$ \sum_{i=0}^{m-1} d_i \;=\; K \cdot M \qquad\text{for some integer } K \ge 0. $$

In the sample, $3 + 4 + 3 = 10 = 2 \cdot 5$, so $K = 2$.

5. Signed representation and why $K$ entries must be “flipped”

Each $d_i$ is a residue mod $M$ and can represent one of two signed integers:

keep positive: signed value $+d_i$;
flip negative: signed value $-(M - d_i)$.

Unit-transfers preserve the signed sum, so to land on the zero vector we need the chosen signed values to sum to $0$.

If we flip $f$ entries (turning each $d_i$ into $-(M - d_i)$), the signed sum equals

$$ \sum d_i - f \cdot M \;=\; (K - f)\cdot M, $$

which is $0$ iff $f = K$. So we must flip exactly $K$ entries.

6. Minimum number of transfers = sum of unflipped entries

Think of the problem as transporting units between piles. Each unflipped pile supplies $d_i$ units; each flipped pile demands $M - d_i$ units. Both sides have the same total (that is why $f = K$). Minimum unit-transfers equals the total supply (equivalently, total demand):

$$ \text{transfers} \;=\; \sum_{\text{unflipped}} d_i. $$

We want this as small as possible, so we should flip the $K$ largest entries of $d$ — i.e. keep the $m - K$ smallest. The answer is

$$ \boxed{\text{answer} \;=\; d_{(0)} + d_{(1)} + \cdots + d_{(m - K - 1)}}, $$

where $d_{(0)} \le d_{(1)} \le \cdots$ is $d$ sorted ascending.

Sample check

Our $d = [3, 4, 3]$, $K = 2$, $m - K = 1$. Sorted: $[3, 3, 4]$. Sum of smallest $1$ entry is $3$, which is the expected answer.

For sample 3, $N = 7$, $M = 12$, $A = [3, 1, 4, 1, 5, 9, 2]$: ignore middle element, pairs give $b = [1, 4, 11]$. Extended $(0, 1, 4, 11, 0)$ gives

$$ d = [1,\; 3,\; 7,\; 1] \pmod{12}. $$

Sum $12$, $K = 1$, $m - K = 3$. Sorted $[1, 1, 3, 7]$, sum of smallest $3$ is $1 + 1 + 3 = 5$. Matches.

How it appears in `model_sol.cpp`

Lines building w: $w = (0,\; b_1,\; \ldots,\; b_h,\; 0)$.
Lines building d: mod-$M$ consecutive differences.
sort(all(d)) sorts $d$ ascending.
k = (int)d.size() - accumulate(all(d), 0LL) / M is exactly $m - K$.
accumulate(d.begin(), d.begin() + k, 0LL) sums the smallest $m - K$ entries.

Complexity and pitfalls

Per test case $O(N \log N)$ due to the sort; fits easily under $\sum N \le 2 \cdot 10^5$.
Use $64$-bit integers for $\sum d$: individual $d_i < M \le 10^9$ and $m$ can be up to $10^5$.
Watch the edge case $N = 1$: no pairs, $b$ is empty, $d = [0]$, and the answer is $0$.