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

<p>In one operation, you can choose any index <code>i</code> where <code>0 &lt;= i &lt; nums.length</code> and change <code>nums[i]</code> to <code>nums[i] + x</code> where <code>x</code> is an integer from the range <code>[-k, k]</code>. You can apply this operation <strong>at most once</strong> for each index <code>i</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 applying the mentioned operation at most once for each index in it</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> 0
<strong>Explanation:</strong> Change nums to be [4, 4, 4]. The score is max(nums) - min(nums) = 4 - 4 = 0.
</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>

#pragma GCC target("avx, mmx, sse2, sse3, sse4")

auto disableSync = [](void) noexcept -> int
{
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    std::cout.tie(nullptr);
    return 0;
}();

class Solution final
{
public:
    int smallestRangeI(const std::vector<int>& nums, const int k) const noexcept
    {
        const auto [itMin, itMax] = std::minmax_element(nums.begin(), nums.end());
        return (*itMin + k < *itMax - k) ? *itMax - *itMin - 2 * k : 0;
    }
};
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