leetcode-problemset/leetcode-cn/originData/find-k-th-smallest-pair-distance.json

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            "questionId": "719",
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            "title": "Find K-th Smallest Pair Distance",
            "titleSlug": "find-k-th-smallest-pair-distance",
            "content": "<p>The <strong>distance of a pair</strong> of integers <code>a</code> and <code>b</code> is defined as the absolute difference between <code>a</code> and <code>b</code>.</p>\n\n<p>Given an integer array <code>nums</code> and an integer <code>k</code>, return <em>the</em> <code>k<sup>th</sup></code> <em>smallest <strong>distance among all the pairs</strong></em> <code>nums[i]</code> <em>and</em> <code>nums[j]</code> <em>where</em> <code>0 &lt;= i &lt; j &lt; nums.length</code>.</p>\n\n<p>&nbsp;</p>\n<p><strong>Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [1,3,1], k = 1\n<strong>Output:</strong> 0\n<strong>Explanation:</strong> Here are all the pairs:\n(1,3) -&gt; 2\n(1,1) -&gt; 0\n(3,1) -&gt; 2\nThen the 1<sup>st</sup> smallest distance pair is (1,1), and its distance is 0.\n</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [1,1,1], k = 2\n<strong>Output:</strong> 0\n</pre>\n\n<p><strong>Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [1,6,1], k = 3\n<strong>Output:</strong> 5\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>n == nums.length</code></li>\n\t<li><code>2 &lt;= n &lt;= 10<sup>4</sup></code></li>\n\t<li><code>0 &lt;= nums[i] &lt;= 10<sup>6</sup></code></li>\n\t<li><code>1 &lt;= k &lt;= n * (n - 1) / 2</code></li>\n</ul>\n",
            "translatedTitle": "找出第 k 小的距离对",
            "translatedContent": "<p>给定一个整数数组，返回所有数对之间的第 k 个最小<strong>距离</strong>。一对 (A, B) 的距离被定义为 A 和 B 之间的绝对差值。</p>\n\n<p><strong>示例 1:</strong></p>\n\n<pre>\n<strong>输入：</strong>\nnums = [1,3,1]\nk = 1\n<strong>输出：0</strong> \n<strong>解释：</strong>\n所有数对如下：\n(1,3) -&gt; 2\n(1,1) -&gt; 0\n(3,1) -&gt; 2\n因此第 1 个最小距离的数对是 (1,1)，它们之间的距离为 0。\n</pre>\n\n<p><strong>提示:</strong></p>\n\n<ol>\n\t<li><code>2 &lt;= len(nums) &lt;= 10000</code>.</li>\n\t<li><code>0 &lt;= nums[i] &lt; 1000000</code>.</li>\n\t<li><code>1 &lt;= k &lt;= len(nums) * (len(nums) - 1) / 2</code>.</li>\n</ol>\n",
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