time limit per test
1 second
memory limit per test
256 megabytes
You are given a string ss, consisting of lowercase English letters. In one operation, you are allowed to swap any two characters of the string ss.
A string ss of length nn is called an anti-palindrome, if si≠sn−i+1si≠sn−i+1 for every ii (1≤i≤n1≤i≤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 ss an anti-palindrome, or output −1−1, if this is not possible.
Input
The first line contains a single integer tt (1≤t≤1041≤t≤104) --- 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 nn (1≤n≤2⋅1051≤n≤2⋅105) --- the length of the string ss.
The second line contains the string ss, consisting of nn lowercase English letters.
The sum of nn over all test cases does not exceed 2⋅1052⋅105.
Output
For each test case, output a single integer --- the minimum number of operations required to make the string ss an anti-palindrome, or −1−1 if this is not possible.
Example
Input
Copy
10
10
codeforces
3
abc
10
taarrrataa
10
dcbdbdcccc
4
wwww
12
cabbaccabaac
10
aadaaaaddc
14
aacdaaaacadcdc
6
abccba
12
dcbcaebacccd
Output
Copy
0
-1
1
1
-1
3
-1
2
2
2Note
In the first test case, the string "codeforces" is already an anti-palindrome, so the answer is 00.
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 −1−1.
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 11.
每项测试的时间限制1秒
每项测试的内存限制256兆字节
给你一个由小写英文字母组成的字符串s
。在一次操作中，你可以交换字符串s
中的任意两个字符。
如果对于每个i（1≤i≤n），si≠sn−i+1
，则长度为n的字符串s
称为反回文。例如，字符串"codeforces"、"string"是反回文，但字符串"abacaba"、"abc"、"test"不是。
确定使字符串s
成为反回文所需的最少操作数，如果不可能，则输出−1
。
输入
第一行包含一个整数t
（1≤t≤104
）---测试用例的数量。测试用例的描述如下。
每个测试用例由两行组成。第一行包含一个整数 n
(1≤n≤2⋅105
) --- 字符串 s
的长度。
第二行包含字符串 s
，由 n
个小写英文字母组成。
所有测试用例的 n
之和不超过 2⋅105
。
输出
对于每个测试用例，输出一个整数 --- 使字符串 s
成为反回文所需的最少操作数，如果不可能，则输出 −1
。
示例
输入复制
10
10
codeforces
3
abc
10
taarrrataa
10
dcbdbdcccc
4
wwww
12
cabbaccabaac
10
aadaaaaddc
14
aacdaaaacadcdc
6
abccba
12
dcbcaebacccd
输出复制
0
-1
1
1
-1
3
-1
2
2
2
注意
在第一个测试用例中，字符串"codeforces"已经是反回文，因此答案为 0
。
在第二个测试用例中，可以证明字符串"abc"无法通过执行允许的操作转换为反回文，因此答案为 −1
。
在第三个测试用例中，只需交换字符串"taarrrataa"的第二个和第五个字符即可，新字符串"trararataa"将是反回文，因此答案为 1
。
代码：
cpp
#include <iostream>
#include <map>
#include <string>
#include <algorithm>
#include <cmath>
#include <vector>
using namespace std;
long long min_operations_to_anti_palindrome(long long n, const string& s) {
if (n % 2 == 1) return -1;
map<char, long long> freq;
for (char c : s) {
freq[c]++;
}
for (const auto& p : freq) {
if (p.second > n / 2) {
return -1;
}
}
map<char, long long> conflict_pairs;
long long cnt = 0;
for (long long i = 0; i < n / 2; i++) {
if (s[i] == s[n - i - 1]) {
conflict_pairs[s[i]]++;
cnt++;
}
}
if (cnt == 0) return 0;
long long num = 0;
for (const auto& p : conflict_pairs) {
num = max(num, p.second);
}
if (2 * num <= cnt) {
return (cnt + 1) / 2;
}
else {
return num;
}
}
int main() {
long long t;
cin >> t;
vector<long long> results;
while (t--) {
long long n;
string s;
cin >> n >> s;
results.push_back(min_operations_to_anti_palindrome(n, s));
}
for (auto& res : results) {
cout << res << endl;
}
return 0;
}