leetcode-problemset/leetcode/originData/maximum-xor-for-each-query.json

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            "title": "Maximum XOR for Each Query",
            "titleSlug": "maximum-xor-for-each-query",
            "content": "<p>You are given a <strong>sorted</strong> array <code>nums</code> of <code>n</code> non-negative integers and an integer <code>maximumBit</code>. You want to perform the following query <code>n</code> <strong>times</strong>:</p>\n\n<ol>\n\t<li>Find a non-negative integer <code>k &lt; 2<sup>maximumBit</sup></code> such that <code>nums[0] XOR nums[1] XOR ... XOR nums[nums.length-1] XOR k</code> is <strong>maximized</strong>. <code>k</code> is the answer to the <code>i<sup>th</sup></code> query.</li>\n\t<li>Remove the <strong>last </strong>element from the current array <code>nums</code>.</li>\n</ol>\n\n<p>Return <em>an array</em> <code>answer</code><em>, where </em><code>answer[i]</code><em> is the answer to the </em><code>i<sup>th</sup></code><em> query</em>.</p>\n\n<p>&nbsp;</p>\n<p><strong>Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [0,1,1,3], maximumBit = 2\n<strong>Output:</strong> [0,3,2,3]\n<strong>Explanation</strong>: The queries are answered as follows:\n1<sup>st</sup> query: nums = [0,1,1,3], k = 0 since 0 XOR 1 XOR 1 XOR 3 XOR 0 = 3.\n2<sup>nd</sup> query: nums = [0,1,1], k = 3 since 0 XOR 1 XOR 1 XOR 3 = 3.\n3<sup>rd</sup> query: nums = [0,1], k = 2 since 0 XOR 1 XOR 2 = 3.\n4<sup>th</sup> query: nums = [0], k = 3 since 0 XOR 3 = 3.\n</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [2,3,4,7], maximumBit = 3\n<strong>Output:</strong> [5,2,6,5]\n<strong>Explanation</strong>: The queries are answered as follows:\n1<sup>st</sup> query: nums = [2,3,4,7], k = 5 since 2 XOR 3 XOR 4 XOR 7 XOR 5 = 7.\n2<sup>nd</sup> query: nums = [2,3,4], k = 2 since 2 XOR 3 XOR 4 XOR 2 = 7.\n3<sup>rd</sup> query: nums = [2,3], k = 6 since 2 XOR 3 XOR 6 = 7.\n4<sup>th</sup> query: nums = [2], k = 5 since 2 XOR 5 = 7.\n</pre>\n\n<p><strong>Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [0,1,2,2,5,7], maximumBit = 3\n<strong>Output:</strong> [4,3,6,4,6,7]\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>nums.length == n</code></li>\n\t<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>\n\t<li><code>1 &lt;= maximumBit &lt;= 20</code></li>\n\t<li><code>0 &lt;= nums[i] &lt; 2<sup>maximumBit</sup></code></li>\n\t<li><code>nums</code>​​​ is sorted in <strong>ascending</strong> order.</li>\n</ul>\n",
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                    "code": "class Solution(object):\n    def getMaximumXor(self, nums, maximumBit):\n        \"\"\"\n        :type nums: List[int]\n        :type maximumBit: int\n        :rtype: List[int]\n        \"\"\"\n        ",
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                    "code": "class Solution:\n    def getMaximumXor(self, nums: List[int], maximumBit: int) -> List[int]:\n        ",
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                    "code": "\n\n/**\n * Note: The returned array must be malloced, assume caller calls free().\n */\nint* getMaximumXor(int* nums, int numsSize, int maximumBit, int* returnSize){\n\n}",
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                    "code": "public class Solution {\n    public int[] GetMaximumXor(int[] nums, int maximumBit) {\n        \n    }\n}",
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                    "code": "/**\n * @param {number[]} nums\n * @param {number} maximumBit\n * @return {number[]}\n */\nvar getMaximumXor = function(nums, maximumBit) {\n    \n};",
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                    "code": "# @param {Integer[]} nums\n# @param {Integer} maximum_bit\n# @return {Integer[]}\ndef get_maximum_xor(nums, maximum_bit)\n    \nend",
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                    "code": "class Solution {\n    func getMaximumXor(_ nums: [Int], _ maximumBit: Int) -> [Int] {\n        \n    }\n}",
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                    "code": "object Solution {\n    def getMaximumXor(nums: Array[Int], maximumBit: Int): Array[Int] = {\n        \n    }\n}",
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                    "code": "impl Solution {\n    pub fn get_maximum_xor(nums: Vec<i32>, maximum_bit: i32) -> Vec<i32> {\n        \n    }\n}",
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                    "code": "defmodule Solution do\n  @spec get_maximum_xor(nums :: [integer], maximum_bit :: integer) :: [integer]\n  def get_maximum_xor(nums, maximum_bit) do\n\n  end\nend",
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            "hints": [
                "Note that the maximum possible XOR result is always 2^(maximumBit) - 1",
                "So the answer for a prefix is the XOR of that prefix XORed with 2^(maximumBit)-1"
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