Introduction

In this blog post, we will explore how to solve the LeetCode problem "908" 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. In one operation, you can choose any index i where 0 <= i < nums.length and change nums[i] to nums[i] + x where x is an integer from the range [-k, k]. You can apply this operation at most once for each index i. The score of nums is the difference between the maximum and minimum elements in nums. Return the minimum score of nums after applying the mentioned operation at most once for each index in it. 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: 0 Explanation: Change nums to be [4, 4, 4]. The score is max(nums) - min(nums) = 4 - 4 = 0. Constraints: 1 <= nums.length <= 104 0 <= nums[i] <= 104 0 <= k <= 104

Explanation

Here's an efficient solution to the problem, along with an explanation:

• Key Idea: The goal is to minimize the difference between the maximum and minimum elements. We can achieve this by increasing the minimum elements as much as possible (by adding k) and decreasing the maximum elements as much as possible (by subtracting k).
• Algorithm: Find the initial maximum and minimum elements of the array. Calculate the potential new maximum and minimum after applying the operations (+k to the original min, and -k to the original max). Return the difference between the new max and new min, unless the new min exceeds the new max, in which case return 0.

• Optimization: No sorting or complex data structures are needed, keeping the solution very efficient.

• Complexity: O(n) runtime, O(1) storage.

Code

    def smallestRangeI(nums, k):
    """
    Calculates the minimum score of nums after applying the mentioned operation at most once for each index.

    Args:
        nums: An integer array.
        k: An integer.

    Returns:
        The minimum score of nums.
    """

    min_val = nums[0]
    max_val = nums[0]

    for num in nums:
        min_val = min(min_val, num)
        max_val = max(max_val, num)

    new_min = min_val + k
    new_max = max_val - k

    return max(0, new_max - new_min)