Introduction

In this blog post, we will explore how to solve the LeetCode problem "2164" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, 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

You are given a 0-indexed integer array nums. Rearrange the values of nums according to the following rules: Sort the values at odd indices of nums in non-increasing order. For example, if nums = [4,1,2,3] before this step, it becomes [4,3,2,1] after. The values at odd indices 1 and 3 are sorted in non-increasing order. Sort the values at even indices of nums in non-decreasing order. For example, if nums = [4,1,2,3] before this step, it becomes [2,1,4,3] after. The values at even indices 0 and 2 are sorted in non-decreasing order. Return the array formed after rearranging the values of nums. Example 1: Input: nums = [4,1,2,3] Output: [2,3,4,1] Explanation: First, we sort the values present at odd indices (1 and 3) in non-increasing order. So, nums changes from [4,1,2,3] to [4,3,2,1]. Next, we sort the values present at even indices (0 and 2) in non-decreasing order. So, nums changes from [4,1,2,3] to [2,3,4,1]. Thus, the array formed after rearranging the values is [2,3,4,1]. Example 2: Input: nums = [2,1] Output: [2,1] Explanation: Since there is exactly one odd index and one even index, no rearrangement of values takes place. The resultant array formed is [2,1], which is the same as the initial array. Constraints: 1 <= nums.length <= 100 1 <= nums[i] <= 100

Explanation

Here's the solution:

• Separate and Sort: Extract elements at even and odd indices into separate lists. Sort the even-indexed elements in non-decreasing order and the odd-indexed elements in non-increasing order.

• Merge: Merge the sorted even and odd elements back into the original array based on their original indices.

• Runtime Complexity: O(n log n), dominated by the sorting operations. Storage Complexity: O(n), due to the creation of separate lists for even and odd indexed numbers.

Code

    def rearrangeArray(nums):
    """
    Rearranges the values of nums according to the given rules.

    Args:
        nums: A list of integers.

    Returns:
        The array formed after rearranging the values of nums.
    """

    odd_indices = []
    even_indices = []

    for i in range(len(nums)):
        if i % 2 == 0:
            even_indices.append(nums[i])
        else:
            odd_indices.append(nums[i])

    even_indices.sort()
    odd_indices.sort(reverse=True)

    even_idx = 0
    odd_idx = 0
    for i in range(len(nums)):
        if i % 2 == 0:
            nums[i] = even_indices[even_idx]
            even_idx += 1
        else:
            nums[i] = odd_indices[odd_idx]
            odd_idx += 1

    return nums