# Solving Leetcode Interviews in Seconds with AI: Beautiful Array


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "932" 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
	> An array nums of length n is beautiful if:  nums is a permutation of the integers in the range [1, n]. For every 0 <= i < j < n, there is no index k with i < k < j where 2 * nums[k] == nums[i] + nums[j].  Given the integer n, return any beautiful array nums of length n. There will be at least one valid answer for the given n.   Example 1: Input: n = 4 Output: [2,1,4,3] Example 2: Input: n = 5 Output: [3,1,2,5,4]    Constraints:  1 <= n <= 1000  

	# Explanation
	Here's a breakdown of the solution:

*   **Divide and Conquer:** The core idea is to recursively construct a beautiful array. We start with a smaller beautiful array (e.g., `[1]`).
*   **Transformation:** We transform the existing beautiful array into a larger one using the properties of odd and even numbers. If `nums` is a beautiful array, then `2*nums[i] - 1` and `2*nums[i]` are also beautiful arrays, but with doubled size and shifted values. We split the range `[1, n]` into odd and even numbers and recursively construct beautiful arrays for each half. The properties ensure that combining these sub-arrays maintains the beautiful array condition.
*   **Base Case:** The base case for the recursion is when n = 1, where the beautiful array is simply `[1]`.

*   **Runtime Complexity:** O(n), **Storage Complexity:** O(n)

	
	# Code
	```python
	class Solution:
    def beautifulArray(self, n: int) -> list[int]:
        """
        Generates a beautiful array of length n.

        Args:
            n: The length of the beautiful array.

        Returns:
            A beautiful array of length n.
        """

        memo = {}  # Memoization to avoid redundant calculations

        def solve(n):
            if n in memo:
                return memo[n]

            if n == 1:
                return [1]

            odds = solve((n + 1) // 2)
            evens = solve(n // 2)

            result = [2 * x - 1 for x in odds] + [2 * x for x in evens]
            memo[n] = result
            return result
        
        return solve(n)
	```