# 1829. Maximum XOR for Each Query

https://leetcode.com/problems/maximum-xor-for-each-query

## Description

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**:

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. Remove the **last** element from the current array `nums`.

Return _an array_ `answer`_, where_ `answer[i]` _is the answer to the_ `ith` _query_.

**Example 1:**

```
**Input:** nums = [0,1,1,3], maximumBit = 2
**Output:** [0,3,2,3]
**Explanation**: The queries are answered as follows:
1st query: nums = [0,1,1,3], k = 0 since 0 XOR 1 XOR 1 XOR 3 XOR 0 = 3.
2nd query: nums = [0,1,1], k = 3 since 0 XOR 1 XOR 1 XOR 3 = 3.
3rd query: nums = [0,1], k = 2 since 0 XOR 1 XOR 2 = 3.
4th 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:
1st query: nums = [2,3,4,7], k = 5 since 2 XOR 3 XOR 4 XOR 7 XOR 5 = 7.
2nd query: nums = [2,3,4], k = 2 since 2 XOR 3 XOR 4 XOR 2 = 7.
3rd query: nums = [2,3], k = 6 since 2 XOR 3 XOR 6 = 7.
4th 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`
* `1 <= n <= 105`
* `1 <= maximumBit <= 20`
* `0 <= nums[i] < 2maximumBit`
* `nums`​​​ is sorted in **ascending** order.

## ac

```java
```
