Formatted question description: https://leetcode.ca/all/1829.html

# 1829. Maximum XOR for Each Query

## Level

Medium

## 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**:

- Find a non-negative integer
`k < 2^maximumBit`

such that`nums[0] XOR nums[1] XOR ... XOR nums[nums.length-1] XOR k`

is**maximized**.`k`

is the answer to the`i-th`

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

.

Return *an array answer, where answer[i] is the answer to the i-th 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 <= 10^5`

`1 <= maximumBit <= 20`

`0 <= nums[i] < 2^maximumBit`

`nums`

is sorted in**ascending**order.

## Solution

Create an array `prefixXors`

of length `n`

, where `prefixXors[i]`

is the xor from `nums[0]`

to `nums[i]`

. Each time, we need to find the `k`

such that `prefixXors[n - 1 - i] ^ k`

is maximized.

Since `k < 2^maximumBit`

and all elements in `nums`

are less than `2^maximumBit`

, obviously all elements in `maximumXors`

are less than `2^maximumBit`

. Therefore, the maximum possible value for `prefixXors[n - 1 - i] ^ k`

is `(1 << maximumBit) - 1`

. The answer to the `i-th`

query is `perfixXors[n - 1 - i] ^ ((1 << maximumBit) - 1)`

.

Finally, return the array of the answers.

```
class Solution {
public int[] getMaximumXor(int[] nums, int maximumBit) {
int maximum = (1 << maximumBit) - 1;
int n = nums.length;
int[] prefixXors = new int[n];
prefixXors[0] = nums[0];
for (int i = 1; i < n; i++)
prefixXors[i] = prefixXors[i - 1] ^ nums[i];
int[] maximumXors = new int[n];
for (int i = 0; i < n; i++) {
int prefixXor = prefixXors[n - 1 - i];
maximumXors[i] = maximum ^ (maximum & prefixXor);
}
return maximumXors;
}
}
```