存量题库数据更新

leetcode-cn/originData/maximum-strong-pair-xor-ii.json

@@ -2,7 +2,7 @@
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             "questionId": "3197",
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             "title": "Maximum Strong Pair XOR II",
@@ -12,13 +12,44 @@
             "translatedContent": "<p>给你一个下标从 <strong>0</strong> 开始的整数数组 <code>nums</code> 。如果一对整数 <code>x</code> 和 <code>y</code> 满足以下条件，则称其为 <strong>强数对</strong> ：</p>\n\n<ul>\n\t<li><code>|x - y| &lt;= min(x, y)</code></li>\n</ul>\n\n<p>你需要从 <code>nums</code> 中选出两个整数，且满足：这两个整数可以形成一个强数对，并且它们的按位异或（<code>XOR</code>）值是在该数组所有强数对中的<strong> 最大值 </strong>。</p>\n\n<p>返回数组 <code>nums</code> 所有可能的强数对中的<strong> 最大 </strong>异或值。</p>\n\n<p><strong>注意</strong>，你可以选择同一个整数两次来形成一个强数对。</p>\n\n<p>&nbsp;</p>\n\n<p><strong class=\"example\">示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [1,2,3,4,5]\n<strong>输出：</strong>7\n<strong>解释：</strong>数组<code> nums </code>中有 11 个强数对：(1, 1), (1, 2), (2, 2), (2, 3), (2, 4), (3, 3), (3, 4), (3, 5), (4, 4), (4, 5) 和 (5, 5) 。\n这些强数对中的最大异或值是 3 XOR 4 = 7 。\n</pre>\n\n<p><strong class=\"example\">示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [10,100]\n<strong>输出：</strong>0\n<strong>解释：</strong>数组<code> nums </code>中有 2 个强数对：(10, 10) 和 (100, 100) 。\n这些强数对中的最大异或值是 10 XOR 10 = 0 ，数对 (100, 100) 的异或值也是 100 XOR 100 = 0 。\n</pre>\n\n<p><strong class=\"example\">示例 3：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [500,520,2500,3000]\n<strong>输出：</strong>1020\n<strong>解释：</strong>数组<code> nums </code>中有 6 个强数对：(500, 500), (500, 520), (520, 520), (2500, 2500), (2500, 3000) 和 (3000, 3000) 。\n这些强数对中的最大异或值是 500 XOR 520 = 1020 ；另一个异或值非零的数对是 (5, 6) ，其异或值是 2500 XOR 3000 = 636 。\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;= 5 * 10<sup>4</sup></code></li>\n\t<li><code>1 &lt;= nums[i] &lt;= 2<sup>20</sup> - 1</code></li>\n</ul>\n",
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             "hints": [
                 "Sort the array, now let <code>x <= y</code> which means <code>|x - y| <= min(x, y)</code> can now be written as <code>y - x <= x</code> or in other words, <code>y <= 2 * x</code>.",
                 "If <code>x</code> and <code>y</code> have the same number of bits, try making<code>y</code>’s bits different from x if possible for each bit starting from the second most significant bit.",