LeetCode 2193: Minimum Number of Moves to Make Palindrome Solution

Master LeetCode problem 2193 (Minimum Number of Moves to Make Palindrome), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2193. Minimum Number of Moves to Make Palindrome

Hard

Two Pointers String Greedy Binary Indexed Tree

Problem Explanation

Explanation:

To solve this problem, we can use a two-pointer approach. We start from the two ends of the string and move towards the center. At each step, we check if the characters at the two pointers are equal. If they are not equal, we count this as a move needed to make the string a palindrome. We continue this process until the two pointers meet in the middle of the string.

The minimum number of moves needed is the sum of the number of moves required at each step. This is because at each step, we are essentially swapping characters to make the string a palindrome.

Algorithm:

Initialize two pointers, one at the start of the string and one at the end.
While the two pointers haven't crossed each other:
- If the characters at the two pointers are different, increment the move count and move the pointers towards the center.
Return the total move count as the minimum number of moves needed to make the string a palindrome.

Time Complexity: The time complexity of this approach is O(n), where n is the length of the input string s.

Space Complexity: The space complexity of this approach is O(1) as we are using constant extra space.

Solution Code

class Solution {
    public int minSwaps(String s) {
        int moves = 0;
        int left = 0, right = s.length() - 1;
        
        while (left < right) {
            if (s.charAt(left) != s.charAt(right)) {
                moves++;
            }
            left++;
            right--;
        }
        
        return moves;
    }
}

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