Smallest Range II
<p>You are given an integer array <code>nums</code> and an integer <code>k</code>.</p>

<p>For each index <code>i</code> where <code>0 &lt;= i &lt; nums.length</code>, change <code>nums[i]</code> to be either <code>nums[i] + k</code> or <code>nums[i] - k</code>.</p>

<p>The <strong>score</strong> of <code>nums</code> is the difference between the maximum and minimum elements in <code>nums</code>.</p>

<p>Return <em>the minimum <strong>score</strong> of </em><code>nums</code><em> after changing the values at each index</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1], k = 0
<strong>Output:</strong> 0
<strong>Explanation:</strong> The score is max(nums) - min(nums) = 1 - 1 = 0.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [0,10], k = 2
<strong>Output:</strong> 6
<strong>Explanation:</strong> Change nums to be [2, 8]. The score is max(nums) - min(nums) = 8 - 2 = 6.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,3,6], k = 3
<strong>Output:</strong> 3
<strong>Explanation:</strong> Change nums to be [4, 6, 3]. The score is max(nums) - min(nums) = 6 - 3 = 3.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>4</sup></code></li>
	<li><code>0 &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
	<li><code>0 &lt;= k &lt;= 10<sup>4</sup></code></li>
</ul>
Greedy; Array; Math; Sorting
class Solution:
    def smallestRangeII(self, nums: List[int], k: int) -> int:
        nums.sort()
        ans = nums[-1] - nums[0]
        for i in range(1, len(nums)):
            mi = min(nums[0] + k, nums[i] - k)
            mx = max(nums[i - 1] + k, nums[-1] - k)
            ans = min(ans, mx - mi)
        return ans
