189. Rotate Array (Approach 1 )

Intitution

To rotate an array to the right by k steps, we essentially want to move each element k positions to the right, with the elements at the end wrapping around to the beginning.

A neat trick is to simulate a circular array:

• If we duplicate the original array, appending it to itself, we create a structure that behaves like an infinite loop of the array.

• Then, to get the rotated version, we can just extract a subarray of length n (original length) starting from the right offset.

This avoids doing complicated index arithmetic or multiple reversals.

Complexity

Space Complexity	Time Complexity
$\text{O}(n)$	$\text{O}(n)$

Code

public void rotate(int[] nums, int k) {
    k = k % nums.length; // Handle cases where k >= nums.length

    if (k == 0) return; // No need to rotate

    // Step 1: Duplicate the array (simulate circular behavior)
    int[] numsDuplicated = new int[2 * nums.length];
    for (int i = 0; i < nums.length; ++i) {
        numsDuplicated[i] = nums[i];
        numsDuplicated[nums.length + i] = nums[i]; // Append the same array
    }

    // Step 2: Copy the rotated portion back into the original array
    for (int i = 0; i < nums.length; ++i) {
        nums[i] = numsDuplicated[nums.length - k + i];
    }
}