leetcode-problemset/leetcode-cn/originData/k-th-smallest-prime-fraction.json

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            "title": "K-th Smallest Prime Fraction",
            "titleSlug": "k-th-smallest-prime-fraction",
            "content": "<p>You are given a sorted integer array <code>arr</code> containing <code>1</code> and <strong>prime</strong> numbers, where all the integers of <code>arr</code> are unique. You are also given an integer <code>k</code>.</p>\n\n<p>For every <code>i</code> and <code>j</code> where <code>0 &lt;= i &lt; j &lt; arr.length</code>, we consider the fraction <code>arr[i] / arr[j]</code>.</p>\n\n<p>Return <em>the</em> <code>k<sup>th</sup></code> <em>smallest fraction considered</em>. Return your answer as an array of integers of size <code>2</code>, where <code>answer[0] == arr[i]</code> and <code>answer[1] == arr[j]</code>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> arr = [1,2,3,5], k = 3\n<strong>Output:</strong> [2,5]\n<strong>Explanation:</strong> The fractions to be considered in sorted order are:\n1/5, 1/3, 2/5, 1/2, 3/5, and 2/3.\nThe third fraction is 2/5.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> arr = [1,7], k = 1\n<strong>Output:</strong> [1,7]\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>2 &lt;= arr.length &lt;= 1000</code></li>\n\t<li><code>1 &lt;= arr[i] &lt;= 3 * 10<sup>4</sup></code></li>\n\t<li><code>arr[0] == 1</code></li>\n\t<li><code>arr[i]</code> is a <strong>prime</strong> number for <code>i &gt; 0</code>.</li>\n\t<li>All the numbers of <code>arr</code> are <strong>unique</strong> and sorted in <strong>strictly increasing</strong> order.</li>\n\t<li><code>1 &lt;= k &lt;= arr.length * (arr.length - 1) / 2</code></li>\n</ul>\n\n<p>&nbsp;</p>\n<strong>Follow up:</strong> Can you solve the problem with better than <code>O(n<sup>2</sup>)</code> complexity?",
            "translatedTitle": "第 K 个最小的素数分数",
            "translatedContent": "<p>给你一个按递增顺序排序的数组 <code>arr</code> 和一个整数 <code>k</code> 。数组 <code>arr</code> 由 <code>1</code> 和若干 <strong>素数</strong>&nbsp; 组成，且其中所有整数互不相同。</p>\n\n<p>对于每对满足 <code>0 &lt;= i &lt; j &lt; arr.length</code> 的 <code>i</code> 和 <code>j</code> ，可以得到分数 <code>arr[i] / arr[j]</code> 。</p>\n\n<p>那么第&nbsp;<code>k</code>&nbsp;个最小的分数是多少呢?&nbsp; 以长度为 <code>2</code> 的整数数组返回你的答案, 这里&nbsp;<code>answer[0] == arr[i]</code>&nbsp;且&nbsp;<code>answer[1] == arr[j]</code> 。</p>\n&nbsp;\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>arr = [1,2,3,5], k = 3\n<strong>输出：</strong>[2,5]\n<strong>解释：</strong>已构造好的分数,排序后如下所示: \n1/5, 1/3, 2/5, 1/2, 3/5, 2/3\n很明显第三个最小的分数是 2/5\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>arr = [1,7], k = 1\n<strong>输出：</strong>[1,7]\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>2 &lt;= arr.length &lt;= 1000</code></li>\n\t<li><code>1 &lt;= arr[i] &lt;= 3 * 10<sup>4</sup></code></li>\n\t<li><code>arr[0] == 1</code></li>\n\t<li><code>arr[i]</code> 是一个 <strong>素数</strong> ，<code>i &gt; 0</code></li>\n\t<li><code>arr</code> 中的所有数字 <strong>互不相同</strong> ，且按 <strong>严格递增</strong> 排序</li>\n\t<li><code>1 &lt;= k &lt;= arr.length * (arr.length - 1) / 2</code></li>\n</ul>\n\n<p>&nbsp;</p>\n\n<p><strong>进阶：</strong>你可以设计并实现时间复杂度小于 <code>O(n<sup>2</sup>)</code> 的算法解决此问题吗？</p>\n",
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