leetcode-problemset/leetcode-cn/originData/maximum-xor-after-operations.json

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            "title": "Maximum XOR After Operations ",
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            "content": "<p>You are given a <strong>0-indexed</strong> integer array <code>nums</code>. In one operation, select <strong>any</strong> non-negative integer <code>x</code> and an index <code>i</code>, then <strong>update</strong> <code>nums[i]</code> to be equal to <code>nums[i] AND (nums[i] XOR x)</code>.</p>\n\n<p>Note that <code>AND</code> is the bitwise AND operation and <code>XOR</code> is the bitwise XOR operation.</p>\n\n<p>Return <em>the <strong>maximum</strong> possible bitwise XOR of all elements of </em><code>nums</code><em> after applying the operation <strong>any number</strong> of times</em>.</p>\n\n<p>&nbsp;</p>\n<p><strong>Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [3,2,4,6]\n<strong>Output:</strong> 7\n<strong>Explanation:</strong> Apply the operation with x = 4 and i = 3, num[3] = 6 AND (6 XOR 4) = 6 AND 2 = 2.\nNow, nums = [3, 2, 4, 2] and the bitwise XOR of all the elements = 3 XOR 2 XOR 4 XOR 2 = 7.\nIt can be shown that 7 is the maximum possible bitwise XOR.\nNote that other operations may be used to achieve a bitwise XOR of 7.</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [1,2,3,9,2]\n<strong>Output:</strong> 11\n<strong>Explanation:</strong> Apply the operation zero times.\nThe bitwise XOR of all the elements = 1 XOR 2 XOR 3 XOR 9 XOR 2 = 11.\nIt can be shown that 11 is the maximum possible bitwise XOR.</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>8</sup></code></li>\n</ul>\n",
            "translatedTitle": "操作后的最大异或和",
            "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;。一次操作中，选择 <strong>任意</strong>&nbsp;非负整数&nbsp;<code>x</code>&nbsp;和一个下标&nbsp;<code>i</code>&nbsp;，<strong>更新</strong>&nbsp;<code>nums[i]</code>&nbsp;为&nbsp;<code>nums[i] AND (nums[i] XOR x)</code>&nbsp;。</p>\n\n<p>注意，<code>AND</code>&nbsp;是逐位与运算，<code>XOR</code>&nbsp;是逐位异或运算。</p>\n\n<p>请你执行 <strong>任意次</strong>&nbsp;更新操作，并返回&nbsp;<code>nums</code>&nbsp;中所有元素&nbsp;<strong>最大</strong>&nbsp;逐位异或和。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre><b>输入：</b>nums = [3,2,4,6]\n<b>输出：</b>7\n<b>解释：</b>选择 x = 4 和 i = 3 进行操作，num[3] = 6 AND (6 XOR 4) = 6 AND 2 = 2 。\n现在，nums = [3, 2, 4, 2] 且所有元素逐位异或得到 3 XOR 2 XOR 4 XOR 2 = 7 。\n可知 7 是能得到的最大逐位异或和。\n注意，其他操作可能也能得到逐位异或和 7 。</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre><b>输入：</b>nums = [1,2,3,9,2]\n<b>输出：</b>11\n<b>解释：</b>执行 0 次操作。\n所有元素的逐位异或和为 1 XOR 2 XOR 3 XOR 9 XOR 2 = 11 。\n可知 11 是能得到的最大逐位异或和。</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>8</sup></code></li>\n</ul>\n",
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                "Consider what it means to be able to choose any x for the operation and which integers could be obtained from a given nums[i].",
                "The given operation can unset any bit in nums[i].",
                "The nth bit of the XOR of all the elements is 1 if the nth bit is 1 for an odd number of elements. When can we ensure it is odd?",
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