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| 1 | +<h2><a href="https://leetcode.com/problems/minimize-the-maximum-difference-of-pairs/?envType=daily-question&envId=2025年06月13日">2616. Minimize the Maximum Difference of Pairs</a></h2><h3>Medium</h3><hr><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> |
| 2 | + |
| 3 | +<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> |
| 4 | + |
| 5 | +<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> |
| 6 | + |
| 7 | +<p> </p> |
| 8 | +<p><strong class="example">Example 1:</strong></p> |
| 9 | + |
| 10 | +<pre> |
| 11 | +<strong>Input:</strong> nums = [10,1,2,7,1,3], p = 2 |
| 12 | +<strong>Output:</strong> 1 |
| 13 | +<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. |
| 14 | +The maximum difference is max(|nums[1] - nums[4]|, |nums[2] - nums[5]|) = max(0, 1) = 1. Therefore, we return 1. |
| 15 | +</pre> |
| 16 | + |
| 17 | +<p><strong class="example">Example 2:</strong></p> |
| 18 | + |
| 19 | +<pre> |
| 20 | +<strong>Input:</strong> nums = [4,2,1,2], p = 1 |
| 21 | +<strong>Output:</strong> 0 |
| 22 | +<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. |
| 23 | +</pre> |
| 24 | + |
| 25 | +<p> </p> |
| 26 | +<p><strong>Constraints:</strong></p> |
| 27 | + |
| 28 | +<ul> |
| 29 | + <li><code>1 <= nums.length <= 10<sup>5</sup></code></li> |
| 30 | + <li><code>0 <= nums[i] <= 10<sup>9</sup></code></li> |
| 31 | + <li><code>0 <= p <= (nums.length)/2</code></li> |
| 32 | +</ul> |
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