Introduction

In this blog post, we will explore how to solve the LeetCode problem "2567" 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 an integer array nums. The low score of nums is the minimum absolute difference between any two integers. The high score of nums is the maximum absolute difference between any two integers. The score of nums is the sum of the high and low scores. Return the minimum score after changing two elements of nums. Example 1: Input: nums = [1,4,7,8,5] Output: 3 Explanation: Change nums[0] and nums[1] to be 6 so that nums becomes [6,6,7,8,5]. The low score is the minimum absolute difference: |6 - 6| = 0. The high score is the maximum absolute difference: |8 - 5| = 3. The sum of high and low score is 3. Example 2: Input: nums = [1,4,3] Output: 0 Explanation: Change nums[1] and nums[2] to 1 so that nums becomes [1,1,1]. The sum of maximum absolute difference and minimum absolute difference is 0. Constraints: 3 <= nums.length <= 105 1 <= nums[i] <= 109

Explanation

Here's the breakdown of the solution:

• Core Idea: The key is to realize that we want to minimize both the low score (minimum absolute difference) and the high score (maximum absolute difference). Changing two numbers strategically allows us to potentially bring the remaining numbers closer together.
• Strategic Changes: We consider scenarios where we modify the two smallest numbers, the two largest numbers, one smallest and one largest number to the median or other strategic value. Sorting the array makes it easy to identify these candidates. We evaluate three possible scenarios:
  1. Keep the smallest element, and discard the two largest. The array will be [nums[0], nums[0], nums[1], ..., nums[n-3]]. The high score is nums[n-3] - nums[0] and low score is 0 (because we changed one element equal to nums[0].
  2. Keep the largest element, and discard the two smallest. The array will be [nums[2], ..., nums[n-2], nums[n-1], nums[n-1]]. The high score is nums[n-1] - nums[2] and low score is 0 (because we changed one element equal to nums[n-1].
  3. Discard the smallest and the largest elements. The array will be [nums[1], nums[1], nums[2], ..., nums[n-2], nums[n-2]]. The high score is nums[n-2] - nums[1] and low score is 0 (because we changed one element equal to nums[1] and one to nums[n-2]).

• Minimization: We calculate the score (high + low) for each scenario and take the minimum score.

• Runtime Complexity: O(n log n) due to sorting.

• Storage Complexity: O(1) (in-place sort, if allowed, otherwise O(n) for a copy).

Code

    def min_score(nums):
    """
    Calculates the minimum score after changing two elements of the array.

    Args:
        nums: A list of integers.

    Returns:
        The minimum score.
    """

    n = len(nums)
    nums.sort()

    # Case 1: Keep the smallest, discard two largest
    score1 = nums[n - 3] - nums[0]

    # Case 2: Keep the largest, discard two smallest
    score2 = nums[n - 1] - nums[2]

    # Case 3: Discard smallest and largest
    score3 = nums[n - 2] - nums[1]

    return min(score1, score2, score3)