> ## Documentation Index
> Fetch the complete documentation index at: https://leetcode-py.wisl.dev/llms.txt
> Use this file to discover all available pages before exploring further.

> ## Agent Instructions
> leetcode-py is a Python LeetCode practice environment generator with one CLI: lcpy. It is not a service or platform.
> Each problem is a directory under leetcode/ with README.md, solution.py, test_solution.py, helpers.py, and playground.ipynb. lcpy gen creates them from JSON templates bundled with the package.
> Examples are backed by tests; copy them verbatim.

# Maximum XOR for Each Query Python Solution

> Tested Python solution for LeetCode 1829 with 19 pytest cases. Generate a practice environment with lcpy.

LeetCode 1829, [Medium](/catalog/medium). Topics: [Array](/catalog/topics/array), [Bit Manipulation](/catalog/topics/bit-manipulation), [Prefix Sum](/catalog/topics/prefix-sum). [View on LeetCode](https://leetcode.com/problems/maximum-xor-for-each-query/description/).

Generate this problem as a practice environment: tested reference solution, 19 [parametrized pytest cases](/practice/testing), and a playground notebook:

```bash theme={"theme":{"light":"github-light","dark":"github-dark"}}
lcpy gen -n 1829   # by problem number
lcpy gen -s maximum_xor_for_each_query   # by problem name
```

## Problem

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 ith query.
* Remove the last element from the current array nums.

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

### Examples

```
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.
```

```
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.
```

```
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

Reference implementation from [solution.py on GitHub](https://github.com/wislertt/leetcode-py/blob/main/leetcode/maximum_xor_for_each_query/solution.py), full suite in [test\_solution.py](https://github.com/wislertt/leetcode-py/blob/main/leetcode/maximum_xor_for_each_query/test_solution.py):

```python theme={"theme":{"light":"github-light","dark":"github-dark"}}
class Solution:
    # Time: O(n)
    # Space: O(1) excluding the output array
    def get_maximum_xor(self, nums: list[int], maximum_bit: int) -> list[int]:
        mask = (1 << maximum_bit) - 1
        total = 0
        for num in nums:
            total ^= num

        answer: list[int] = []
        for num in reversed(nums):
            answer.append(total ^ mask)
            total ^= num
        return answer
```

## Complexity

| Time | Space |
| - | - |
| O(n) | O(1) excluding the output array |

## Tags

[NeetCode All](/catalog/neetcode).


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