1829 - Maximum XOR for Each Query.

Medium

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:

• Find a non-negative integer <code>k < 2^maximumBit</code> such that nums[0] XOR nums[1] XOR ... XOR nums[nums.length-1] XOR k is maximized. k is the answer to the <code>i^th</code> query.

• Remove the last element from the current array nums.

Return an array answer, where answer[i] is the answer to the <code>i^th</code> query.

Example 1:

Input: nums = 0,1,1,3, maximumBit = 2

Output: 0,3,2,3

Explanation: The queries are answered as follows:

1^st query: nums = 0,1,1,3, k = 0 since 0 XOR 1 XOR 1 XOR 3 XOR 0 = 3.

2^nd query: nums = 0,1,1, k = 3 since 0 XOR 1 XOR 1 XOR 3 = 3.

3^rd query: nums = 0,1, k = 2 since 0 XOR 1 XOR 2 = 3.

4^th query: nums = 0, k = 3 since 0 XOR 3 = 3.

Example 2:

Input: nums = 2,3,4,7, maximumBit = 3

Output: 5,2,6,5

Explanation: The queries are answered as follows:

1^st query: nums = 2,3,4,7, k = 5 since 2 XOR 3 XOR 4 XOR 7 XOR 5 = 7.

2^nd query: nums = 2,3,4, k = 2 since 2 XOR 3 XOR 4 XOR 2 = 7.

3^rd query: nums = 2,3, k = 6 since 2 XOR 3 XOR 6 = 7.

4^th query: nums = 2, k = 5 since 2 XOR 5 = 7.

Example 3:

Input: nums = 0,1,2,2,5,7, maximumBit = 3

Output: 4,3,6,4,6,7

Constraints:

• nums.length == n

• <code>1 <= n <= 10⁵</code>

• 1 <= maximumBit <= 20

• <code>0 <= numsi< 2^maximumBit</code>

• nums is sorted in ascending order.