leetcode-problemset/leetcode-cn/originData/minimize-the-maximum-difference-of-pairs.json

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            "title": "Minimize the Maximum Difference of Pairs",
            "titleSlug": "minimize-the-maximum-difference-of-pairs",
            "content": "<p>You are given a <strong>0-indexed</strong> integer array <code>nums</code> and an integer <code>p</code>. Find <code>p</code> pairs of indices of <code>nums</code> such that the <strong>maximum</strong> difference amongst all the pairs is <strong>minimized</strong>. Also, ensure no index appears more than once amongst the <code>p</code> pairs.</p>\n\n<p>Note that for a pair of elements at the index <code>i</code> and <code>j</code>, the difference of this pair is <code>|nums[i] - nums[j]|</code>, where <code>|x|</code> represents the <strong>absolute</strong> <strong>value</strong> of <code>x</code>.</p>\n\n<p>Return <em>the <strong>minimum</strong> <strong>maximum</strong> difference among all </em><code>p</code> <em>pairs.</em> We define the maximum of an empty set to be zero.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [10,1,2,7,1,3], p = 2\n<strong>Output:</strong> 1\n<strong>Explanation:</strong> The first pair is formed from the indices 1 and 4, and the second pair is formed from the indices 2 and 5. \nThe maximum difference is max(|nums[1] - nums[4]|, |nums[2] - nums[5]|) = max(0, 1) = 1. Therefore, we return 1.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [4,2,1,2], p = 1\n<strong>Output:</strong> 0\n<strong>Explanation:</strong> Let the indices 1 and 3 form a pair. The difference of that pair is |2 - 2| = 0, which is the minimum we can attain.\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>5</sup></code></li>\n\t<li><code>0 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>\n\t<li><code>0 &lt;= p &lt;= (nums.length)/2</code></li>\n</ul>\n",
            "translatedTitle": "最小化数对的最大差值",
            "translatedContent": "<p>给你一个下标从 <strong>0</strong>&nbsp;开始的整数数组&nbsp;<code>nums</code>&nbsp;和一个整数&nbsp;<code>p</code>&nbsp;。请你从&nbsp;<code>nums</code>&nbsp;中找到&nbsp;<code>p</code> 个下标对，每个下标对对应数值取差值，你需要使得这 <code>p</code> 个差值的&nbsp;<strong>最大值</strong>&nbsp;<strong>最小</strong>。同时，你需要确保每个下标在这&nbsp;<code>p</code>&nbsp;个下标对中最多出现一次。</p>\n\n<p>对于一个下标对&nbsp;<code>i</code>&nbsp;和&nbsp;<code>j</code>&nbsp;，这一对的差值为&nbsp;<code>|nums[i] - nums[j]|</code>&nbsp;，其中&nbsp;<code>|x|</code>&nbsp;表示 <code>x</code>&nbsp;的 <strong>绝对值</strong>&nbsp;。</p>\n\n<p>请你返回 <code>p</code>&nbsp;个下标对对应数值 <strong>最大差值</strong>&nbsp;的 <strong>最小值</strong>&nbsp;。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<b>输入：</b>nums = [10,1,2,7,1,3], p = 2\n<b>输出：</b>1\n<b>解释：</b>第一个下标对选择 1 和 4 ，第二个下标对选择 2 和 5 。\n最大差值为 max(|nums[1] - nums[4]|, |nums[2] - nums[5]|) = max(0, 1) = 1 。所以我们返回 1 。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<b>输入：</b>nums = [4,2,1,2], p = 1\n<b>输出：</b>0\n<b>解释：</b>选择下标 1 和 3 构成下标对。差值为 |2 - 2| = 0 ，这是最大差值的最小值。\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>5</sup></code></li>\n\t<li><code>0 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>\n\t<li><code>0 &lt;= p &lt;= (nums.length)/2</code></li>\n</ul>\n",
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            "hints": [
                "Can we use dynamic programming here?",
                "To minimize the answer, the array should be sorted first.",
                "The recurrence relation is fn(i, x) = min(fn(i+1, x), max(abs(nums[i]-nums[i+1]), fn(i+2, p-1)), and fn(0,p) gives the desired answer."
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