Minimum swaps required in the String such that S[i] != S[n-i+1]

Given a string S consisting of lowercase characters, the task is to find the minimum number of swaps required in the string such that S[i] != S[n-i+1], where 1 <= i <= n.

Examples:

Input: N = 3, S = "ABC"
Output: -1
Explanation: In the string "abc" , s[2] = s[3-2+1] even after applying the allowed operation.

Input: N = 10, S = "taarrrataa"
Output: 1
Explanation: It is enough to swap the 2nd and 5th characters of the string "taarrrataa" and the new string "trararataa" satisfies the given condition.

Approach: To solve the problem follow the below idea:

If the string has an odd number of letters, you can't make any solution. This is because there will always be a middle letter that can't be paired with anything else. If the string has an even number of letters, you have to do some operations

Below are steps to solve the problem:

• Two vectors p and x are initialized, each with 26 elements, to keep track of symmetric pairs and character occurrences.
• Iterate through the first half of the input String:
  • If a character is the same at symmetric positions within the string, it increments the count for that character's symmetric pair (p vector).
  • It also increments the overall count of that character (x vector).
  • Then check if any character appears more than half the total characters (N/2). If so, it's not possible to make an anti-palindrome, and -1 is printed.
• Calculate the total count of symmetric pairs (count) and the maximum count of symmetric pairs (maxi).

Below is the implementation of the above approach:

1

Time Complexity: O(N)
Auxiliary Space: O(1)