[Binary Search]Rotate List Left by K

https://binarysearch.com/problems/Rotate-List-Left-by-K

Write a function that rotates a list of numbers to the left by k elements. Numbers should wrap around.

Note: The list is guaranteed to have at least one element, and k is guaranteed to be less than or equal to the length of the list.

Bonus: Do this without creating a copy of the list. How many swap or move operations do you need?

Constraints

n ≤ 100,000 where n is the length of nums

Example 1

Input:

nums = [1, 2, 3, 4, 5, 6]
k = 2

Output: [3, 4, 5, 6, 1, 2]

Example 2

Input

nums = [1, 2, 3, 4, 5, 6]
k = 6

Output: [1, 2, 3, 4, 5, 6]

Example 3

Input

nums = [1]
k = 0

Output: [1]

class Solution {
    public int[] solve(int[] nums, int k) {
        rotate(nums, 0, k - 1);
        rotate(nums, k, nums.length - 1);
        rotate(nums, 0, nums.length - 1);
        return nums;
    }

    private void rotate(int[] nums, int left, int right) {
        while (left < right) {
            int temp = nums[left];
            nums[left] = nums[right];
            nums[right] = temp;
            left++;
            right--;
        }
    }
}