leetcode-problemset/leetcode-cn/originData/smallest-range-i.json

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            "title": "Smallest Range I",
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            "content": "<p>You are given an integer array <code>nums</code> and an integer <code>k</code>.</p>\n\n<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>\n\n<p>The <strong>score</strong> of <code>nums</code> is the difference between the maximum and minimum elements in <code>nums</code>.</p>\n\n<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>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [1], k = 0\n<strong>Output:</strong> 0\n<strong>Explanation:</strong> The score is max(nums) - min(nums) = 1 - 1 = 0.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [0,10], k = 2\n<strong>Output:</strong> 6\n<strong>Explanation:</strong> Change nums to be [2, 8]. The score is max(nums) - min(nums) = 8 - 2 = 6.\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [1,3,6], k = 3\n<strong>Output:</strong> 0\n<strong>Explanation:</strong> Change nums to be [4, 4, 4]. The score is max(nums) - min(nums) = 4 - 4 = 0.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= nums.length &lt;= 10<sup>4</sup></code></li>\n\t<li><code>0 &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>\n\t<li><code>0 &lt;= k &lt;= 10<sup>4</sup></code></li>\n</ul>\n",
            "translatedTitle": "最小差值 I",
            "translatedContent": "<p>给你一个整数数组 <code>nums</code>，和一个整数 <code>k</code> 。</p>\n\n<p>在一个操作中，您可以选择 <code>0 &lt;= i &lt; nums.length</code> 的任何索引 <code>i</code> 。将 <code>nums[i]</code> 改为 <code>nums[i] + x</code> ，其中 <code>x</code> 是一个范围为 <code>[-k, k]</code> 的整数。对于每个索引 <code>i</code> ，最多 <strong>只能 </strong>应用 <strong>一次</strong> 此操作。</p>\n\n<p><code>nums</code>&nbsp;的&nbsp;<strong>分数&nbsp;</strong>是&nbsp;<code>nums</code>&nbsp;中最大和最小元素的差值。&nbsp;</p>\n\n<p><em>在对&nbsp; <code>nums</code> 中的每个索引最多应用一次上述操作后，返回&nbsp;<code>nums</code> 的最低 <strong>分数</strong></em> 。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [1], k = 0\n<strong>输出：</strong>0\n<strong>解释：</strong>分数是 max(nums) - min(nums) = 1 - 1 = 0。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [0,10], k = 2\n<strong>输出：</strong>6\n<strong>解释：</strong>将 nums 改为 [2,8]。分数是 max(nums) - min(nums) = 8 - 2 = 6。\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [1,3,6], k = 3\n<strong>输出：</strong>0\n<strong>解释：</strong>将 nums 改为 [4,4,4]。分数是 max(nums) - min(nums) = 4 - 4 = 0。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= nums.length &lt;= 10<sup>4</sup></code></li>\n\t<li><code>0 &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>\n\t<li><code>0 &lt;= k &lt;= 10<sup>4</sup></code></li>\n</ul>\n",
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