541. Reverse String II | zhengxingxing

Method 1

Given a string s and an integer k, reverse the first k characters for every 2k characters counting from the start of the string.
If there are fewer than k characters left, reverse all of them. If there are less than 2k but greater than or equal to k characters, then reverse the first k characters and leave the other as original.

Example 1

Input: s = "abcdefg", k = 2
Output: "bacdfeg"

Example 2

Input: s = "abcd", k = 2
Output: "bacd"

Method 1

【O(n) time | O(n) space】

package Leetcode.TwoPointer;

/**
 * @author zhengxingxing
 * @date 2025/01/31
 */
public class ReverseStringII {
    public static String reverseStr(String s, int k) {
        char[] arr = s.toCharArray();

        for (int i = 0; i < arr.length; i += 2 * k) {
            // Calculate the actual length that needs to be reversed each time
            int left = i;
            // Use Math.min to handle boundary cases when remaining chars less than k
            int right = Math.min(i + k - 1, arr.length - 1);

            // Two pointers approach to reverse characters
            while (left < right){
                char temp = arr[left];
                arr[left] = arr[right];
                arr[right] = temp;
                left++;
                right--;
            }
        }
        return new String(arr);
    }

    public static void main(String[] args) {
        // Test case 1: Regular case
        String s1 = "abcdefg";
        int k1 = 2;
        System.out.println("Test case 1 result: " + reverseStr(s1, k1)); // Expected output: "bacdfeg"

        // Test case 2: String length equals 2k
        String s2 = "abcd";
        int k2 = 2;
        System.out.println("Test case 2 result: " + reverseStr(s2, k2)); // Expected output: "bacd"

        // Test case 3: Edge case - single character
        String s3 = "a";
        int k3 = 2;
        System.out.println("Test case 3 result: " + reverseStr(s3, k3)); // Expected output: "a"
    }
}

Method 1

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