# Solving Leetcode Interviews in Seconds with AI: Rotate Array


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "189" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> Given an integer array nums, rotate the array to the right by k steps, where k is non-negative.   Example 1:  Input: nums = [1,2,3,4,5,6,7], k = 3 Output: [5,6,7,1,2,3,4] Explanation: rotate 1 steps to the right: [7,1,2,3,4,5,6] rotate 2 steps to the right: [6,7,1,2,3,4,5] rotate 3 steps to the right: [5,6,7,1,2,3,4]  Example 2:  Input: nums = [-1,-100,3,99], k = 2 Output: [3,99,-1,-100] Explanation:  rotate 1 steps to the right: [99,-1,-100,3] rotate 2 steps to the right: [3,99,-1,-100]    Constraints:  1 <= nums.length <= 105 -231 <= nums[i] <= 231 - 1 0 <= k <= 105    Follow up:  Try to come up with as many solutions as you can. There are at least three different ways to solve this problem. Could you do it in-place with O(1) extra space?  

	# Explanation
	Here's a breakdown of the solution, followed by the Python code:

*   **Key Idea:** The most efficient in-place approach involves reversing array segments. First, reverse the entire array. Then, reverse the first `k` elements, and finally, reverse the remaining `n-k` elements. This achieves the rotation in O(1) space. We take `k` modulo `n` to handle cases where `k` is larger than the array length.
*   **In-Place Reversal:** The reversal operation is done in-place using a standard two-pointer approach.

*   **Complexity:** O(n) time complexity, O(1) space complexity.

	
	# Code
	```python
	def rotate(nums, k):
    """
    Rotates an array to the right by k steps in-place.

    Args:
        nums: The list of numbers to rotate.
        k: The number of steps to rotate.
    """
    n = len(nums)
    k %= n  # Handle cases where k > n

    def reverse(arr, start, end):
        """Reverses a portion of the array in-place."""
        while start < end:
            arr[start], arr[end] = arr[end], arr[start]
            start += 1
            end -= 1

    # 1. Reverse the entire array
    reverse(nums, 0, n - 1)

    # 2. Reverse the first k elements
    reverse(nums, 0, k - 1)

    # 3. Reverse the remaining n-k elements
    reverse(nums, k, n - 1)
	```