# Solving Leetcode Interviews in Seconds with AI: Concatenation of Array


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1929" 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 of length n, you want to create an array ans of length 2n where ans[i] == nums[i] and ans[i + n] == nums[i] for 0 <= i < n (0-indexed). Specifically, ans is the concatenation of two nums arrays. Return the array ans.   Example 1:  Input: nums = [1,2,1] Output: [1,2,1,1,2,1] Explanation: The array ans is formed as follows: - ans = [nums[0],nums[1],nums[2],nums[0],nums[1],nums[2]] - ans = [1,2,1,1,2,1] Example 2:  Input: nums = [1,3,2,1] Output: [1,3,2,1,1,3,2,1] Explanation: The array ans is formed as follows: - ans = [nums[0],nums[1],nums[2],nums[3],nums[0],nums[1],nums[2],nums[3]] - ans = [1,3,2,1,1,3,2,1]    Constraints:  n == nums.length 1 <= n <= 1000 1 <= nums[i] <= 1000  

	# Explanation
	*   Create a new array `ans` with a length twice the length of the input array `nums`.
*   Populate the first half of `ans` with the elements of `nums`.
*   Populate the second half of `ans` with the elements of `nums` again, effectively concatenating `nums` with itself.

*   **Runtime Complexity:** O(n), where n is the length of the input array `nums`. **Storage Complexity:** O(n), as we create a new array `ans` of size 2n.

	
	# Code
	```python
	def getConcatenation(nums):
    """
    Given an integer array nums of length n, returns an array ans of length 2n where ans[i] == nums[i] and ans[i + n] == nums[i] for 0 <= i < n (0-indexed).

    Args:
        nums (list[int]): An integer array.

    Returns:
        list[int]: The concatenation of two nums arrays.
    """
    n = len(nums)
    ans = [0] * (2 * n)

    for i in range(n):
        ans[i] = nums[i]
        ans[i + n] = nums[i]

    return ans
	```