Introduction

In this blog post, we will explore how to solve the LeetCode problem "910" 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 and an integer k. For each index i where 0 <= i < nums.length, change nums[i] to be either nums[i] + k or nums[i] - k. The score of nums is the difference between the maximum and minimum elements in nums. Return the minimum score of nums after changing the values at each index. Example 1: Input: nums = [1], k = 0 Output: 0 Explanation: The score is max(nums) - min(nums) = 1 - 1 = 0. Example 2: Input: nums = [0,10], k = 2 Output: 6 Explanation: Change nums to be [2, 8]. The score is max(nums) - min(nums) = 8 - 2 = 6. Example 3: Input: nums = [1,3,6], k = 3 Output: 3 Explanation: Change nums to be [4, 6, 3]. The score is max(nums) - min(nums) = 6 - 3 = 3. Constraints: 1 <= nums.length <= 104 0 <= nums[i] <= 104 0 <= k <= 104

Explanation

Here's the solution to minimize the range after adding or subtracting k from each element:

• Sorting: Sort the input array nums. This allows us to efficiently determine potential minimum and maximum values after the transformation.

• Iterative Optimization: Iterate through the sorted array. For each element, consider the possibilities of it being the boundary between elements that are incremented and decremented. Calculate the potential maximum and minimum values for each possible boundary and keep track of the minimal range found so far.

• Runtime Complexity: O(n log n) due to sorting. Storage Complexity: O(1) excluding the input array.

Code

    def smallestRangeII(nums, k):
    nums.sort()
    n = len(nums)
    ans = nums[-1] - nums[0]

    for i in range(n - 1):
        high = max(nums[-1] - k, nums[i] + k)
        low = min(nums[0] + k, nums[i + 1] - k)
        ans = min(ans, high - low)

    return ans