Given a string s, return the minimum number of moves required to transform s into a palindrome. In each move, you can swap any two adjacent characters in s.

Note: The input string is guaranteed to be convertible into a palindrome.

Constraints:

• $0 \le$ s.length $\le 2000$

• s consists of only lowercase English letters.

• s is guaranteed to be converted into a palindrome in a finite number of moves.

The main strategy for solving this problem is to use a two-pointer approach to progressively match characters from the outer ends of the string toward the center, while minimizing adjacent swaps to transform the string into a palindrome. For each character on the left side, the algorithm searches for its matching counterpart on the right side and moves it into place by repeatedly swapping adjacent characters. If a match is found, the right-side pointer moves inward; if no match is found, it indicates that the character is the center of an odd-length palindrome and is positioned accordingly.

Using the above intuition, the solution can be implemented as follows:

1. Initialize a variable, moves, with $0$ to keep track of the number of swaps required.

2. Initialize two pointers, i at the beginning of the string and j at the end of the string, to traverse the string from both ends toward the center.

   1. At each iteration, the goal is to match the character at position i with the corresponding character at position j.

3. Start an inner loop with k initialized to j, which represents the current character at the end of the string. It moves backward from j to i to find a matching character for s[i].

   1. The loop checks whether s[i] == s[k]. If a match is found, we keep swapping s[k] with s[k+1] until k reaches j. For each swap, increment the moves counter.

   2. After the character is moved to position j, decrement j to continue processing the next character from the end.

4. If no match is found by the time k reaches i (i.e., k == i), it means that the character at i is the center character of an odd-length palindrome.

   1. In this case, the number of moves is incremented by (s.size() / 2) - i, which is the number of moves required to bring this unique character to the center of the string. In this case, we don't swap any characters; just update moves.

5. After processing the entire string, return the value of moves, which represents the minimum number of moves needed to transform the input string into a palindrome.

Let’s look at the following illustration to get a better understanding of the solution: