Making Anti-Palindromes
Contest: Codeforces Round 867 (Div. 3)
Difficulty rating: 1600
Tags: greedy, math, strings

You are given a string $$$s$$$, consisting of lowercase English letters. In one operation, you are allowed to swap any two characters of the string $$$s$$$.

A string $$$s$$$ of length $$$n$$$ is called an anti-palindrome, if $$$s[i] \ne s[n - i + 1]$$$ for every $$$i$$$ ($$$1 \le i \le n$$$). For example, the strings "codeforces", "string" are anti-palindromes, but the strings "abacaba", "abc", "test" are not.

Determine the minimum number of operations required to make the string $$$s$$$ an anti-palindrome, or output $$$-1$$$, if this is not possible.

Input format:
The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. The description of the test cases follows.

Each test case consists of two lines. The first line contains a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the length of the string $$$s$$$.

The second line contains the string $$$s$$$, consisting of $$$n$$$ lowercase English letters.

The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output format:
For each test case, output a single integer — the minimum number of operations required to make the string $$$s$$$ an anti-palindrome, or $$$-1$$$ if this is not possible.

Constraints:
Time limit: 1.0 s
Memory limit: 256.0 MB

Note:
In the first test case, the string "codeforces" is already an anti-palindrome, so the answer is $$$0$$$.

In the second test case, it can be shown that the string "abc" cannot be transformed into an anti-palindrome by performing the allowed operations, so the answer is $$$-1$$$.

In the third test case, it is enough to swap the second and the fifth characters of the string "taarrrataa", and the new string "trararataa" will be an anti-palindrome, so the answer is $$$1$$$.

Example input:
10
10
codeforces
3
abc
10
taarrrataa
10
dcbdbdcccc
4
wwww
12
cabbaccabaac
10
aadaaaaddc
14
aacdaaaacadcdc
6
abccba
12
dcbcaebacccd

Example output:
0
-1
1
1
-1
3
-1
2
2
2

Editorial:
If $$$n$$$ is odd, then there is no solution, since $$$s[(n + 1) / 2] = s[(n + 1) - (n + 1) / 2]$$$.
If $$$n$$$ is even, then all symbols are split into pairs $$$s[i], s[n + 1 - i]$$$. Let's denote the number of occurrences of the symbol $$$c$$$ as $$$cnt[c]$$$.
Note that if $$$cnt[c] > n / 2$$$ for some $$$c$$$, then after applying the operations there will be a pair where both characters are equal to $$$c$$$, then it is impossible to make the string $$$s$$$ anti-palindrome.
Otherwise, we will calculate $$$k$$$ — the number of pairs, where $$$s[i] = s[n + 1 - i]$$$, we will also find $$$m$$$ — the maximum number of pairs, where $$$s[i] = s[n + 1 - i] = c$$$, for all characters $$$c$$$. Let $$$x$$$ be a symbol for which the number of such pairs is equal to $$$m$$$. Note that $$$ans\ge m$$$, because in one operation the number of pairs where $$$s[i] = s[n + 1 - i] = x$$$ cannot decrease by more than $$$1$$$. Also note that $$$ans \ge \lceil \frac{k}{2} \rceil$$$, because for each operation we reduce the number of pairs where $$$s[i] = s[n + 1 - i]$$$ by no more than $$$2$$$. It turns out that $$$ans = max(m, \lceil \frac{k}{2} \rceil)$$$, to show this, you can act greedily — until $$$k > 0$$$:
• If $$$k=m$$$, then we find a pair $$$s[i] = s[n + 1 - i] = x$$$, since $$$cnt[x] \le n / 2$$$, then there is a pair where $$$s[j] \ne x$$$ and $$$s[n + 1 - j] \ne x$$$. Then swap $$$s[i]$$$ and $$$s[j]$$$.
• Otherwise, find the pair $$$s[i] = s[n + 1 - i] = x$$$, and the pair $$$s[j] = s[n + 1 - j] \ne x$$$. Then swap $$$s[i]$$$ and $$$s[j]$$$.
It is not difficult to check that in both cases, $$$max(m, \lceil k/2\rceil)$$$ will decrease by exactly $$$1$$$, which means $$$ans$$$ is achieved with this algorithm.