{ "data": { "question": { "questionId": "1940", "questionFrontendId": "1829", "categoryTitle": "Algorithms", "boundTopicId": 722724, "title": "Maximum XOR for Each Query", "titleSlug": "maximum-xor-for-each-query", "content": "

You are given a sorted array nums of n non-negative integers and an integer maximumBit. You want to perform the following query n times:

\n\n
    \n\t
  1. Find a non-negative integer k < 2maximumBit such that nums[0] XOR nums[1] XOR ... XOR nums[nums.length-1] XOR k is maximized. k is the answer to the ith query.
    2. \n\t
  2. Remove the last element from the current array nums.
    3. \n
\n\n

Return an array answer, where answer[i] is the answer to the ith query.

\n\n

 

\n

Example 1:

\n\n
\nInput: nums = [0,1,1,3], maximumBit = 2\nOutput: [0,3,2,3]\nExplanation: The queries are answered as follows:\n1st query: nums = [0,1,1,3], k = 0 since 0 XOR 1 XOR 1 XOR 3 XOR 0 = 3.\n2nd query: nums = [0,1,1], k = 3 since 0 XOR 1 XOR 1 XOR 3 = 3.\n3rd query: nums = [0,1], k = 2 since 0 XOR 1 XOR 2 = 3.\n4th query: nums = [0], k = 3 since 0 XOR 3 = 3.\n
\n\n

Example 2:

\n\n
\nInput: nums = [2,3,4,7], maximumBit = 3\nOutput: [5,2,6,5]\nExplanation: The queries are answered as follows:\n1st query: nums = [2,3,4,7], k = 5 since 2 XOR 3 XOR 4 XOR 7 XOR 5 = 7.\n2nd query: nums = [2,3,4], k = 2 since 2 XOR 3 XOR 4 XOR 2 = 7.\n3rd query: nums = [2,3], k = 6 since 2 XOR 3 XOR 6 = 7.\n4th query: nums = [2], k = 5 since 2 XOR 5 = 7.\n
\n\n

Example 3:

\n\n
\nInput: nums = [0,1,2,2,5,7], maximumBit = 3\nOutput: [4,3,6,4,6,7]\n
\n\n

 

\n

Constraints:

\n\n\n", "translatedTitle": "每个查询的最大异或值", "translatedContent": "

给你一个 有序 数组 nums ，它由 n 个非负整数组成，同时给你一个整数 maximumBit 。你需要执行以下查询 n 次：

\n\n
    \n\t
  1. 找到一个非负整数 k < 2maximumBit ，使得 nums[0] XOR nums[1] XOR ... XOR nums[nums.length-1] XOR k 的结果 最大化 。k 是第 i 个查询的答案。
    2. \n\t
  2. 从当前数组 nums 删除 最后 一个元素。
    3. \n
\n\n

请你返回一个数组 answer ，其中 answer[i]是第 i 个查询的结果。

\n\n

 

\n\n

示例 1：

\n\n
\n输入：nums = [0,1,1,3], maximumBit = 2\n输出：[0,3,2,3]\n解释：查询的答案如下：\n第一个查询：nums = [0,1,1,3]，k = 0，因为 0 XOR 1 XOR 1 XOR 3 XOR 0 = 3 。\n第二个查询：nums = [0,1,1]，k = 3，因为 0 XOR 1 XOR 1 XOR 3 = 3 。\n第三个查询：nums = [0,1]，k = 2，因为 0 XOR 1 XOR 2 = 3 。\n第四个查询：nums = [0]，k = 3，因为 0 XOR 3 = 3 。\n
\n\n

示例 2：

\n\n
\n输入：nums = [2,3,4,7], maximumBit = 3\n输出：[5,2,6,5]\n解释：查询的答案如下：\n第一个查询：nums = [2,3,4,7]，k = 5，因为 2 XOR 3 XOR 4 XOR 7 XOR 5 = 7。\n第二个查询：nums = [2,3,4]，k = 2，因为 2 XOR 3 XOR 4 XOR 2 = 7 。\n第三个查询：nums = [2,3]，k = 6，因为 2 XOR 3 XOR 6 = 7 。\n第四个查询：nums = [2]，k = 5，因为 2 XOR 5 = 7 。\n
\n\n

示例 3：

\n\n
\n输入：nums = [0,1,2,2,5,7], maximumBit = 3\n输出：[4,3,6,4,6,7]\n
\n\n

 

\n\n

提示：

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