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:
s.length
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:
Initialize a variable, moves, with
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.
At each iteration, the goal is to match the character at position i with the corresponding character at position j.
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].
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.
After the character is moved to position j, decrement j to continue processing the next character from the end.
If no match is found by the time k reaches ...
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:
s.length
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:
Initialize a variable, moves, with
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.
At each iteration, the goal is to match the character at position i with the corresponding character at position j.
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].
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.
After the character is moved to position j, decrement j to continue processing the next character from the end.
If no match is found by the time k reaches ...