# 1829. 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.

## Solutions

Solution 1: Bitwise Operation + Enumeration

First, we preprocess the XOR sum $xs$ of the array nums, i.e., $xs=nums[0] \oplus nums[1] \oplus \cdots \oplus nums[n-1]$.

Next, we enumerate each element $x$ in the array nums from back to front. The current XOR sum is $xs$. We need to find a number $k$ such that the value of $xs \oplus k$ is as large as possible, and $k \lt 2^{maximumBit}$.

That is to say, we start from the $maximumBit - 1$ bit of $xs$ and enumerate to the lower bit. If a bit of $xs$ is $0$, then we set the corresponding bit of $k$ to $1$. Otherwise, we set the corresponding bit of $k$ to $0$. In this way, the final $k$ is the answer to each query. Then, we update $xs$ to $xs \oplus x$ and continue to enumerate the next element.

The time complexity is $O(n \times m)$, where $n$ and $m$ are the values of the array nums and maximumBit respectively. Ignoring the space consumption of the answer, the space complexity is $O(1)$.

Solution 2: Enumeration Optimization

Similar to Solution 1, we first preprocess the XOR sum $xs$ of the array nums, i.e., $xs=nums[0] \oplus nums[1] \oplus \cdots \oplus nums[n-1]$.

Next, we calculate $2^{maximumBit} - 1$, which is $2^{maximumBit}$ minus $1$, denoted as $mask$. Then, we enumerate each element $x$ in the array nums from back to front. The current XOR sum is $xs$, then $k=xs \oplus mask$ is the answer to each query. Then, we update $xs$ to $xs \oplus x$ and continue to enumerate the next element.

The time complexity is $O(n)$, where $n$ is the length of the array nums. Ignoring the space consumption of the answer, the space complexity is $O(1)$.

• class Solution {
public int[] getMaximumXor(int[] nums, int maximumBit) {
int n = nums.length;
int xs = 0;
for (int x : nums) {
xs ^= x;
}
int[] ans = new int[n];
for (int i = 0; i < n; ++i) {
int x = nums[n - i - 1];
int k = 0;
for (int j = maximumBit - 1; j >= 0; --j) {
if (((xs >> j) & 1) == 0) {
k |= 1 << j;
}
}
ans[i] = k;
xs ^= x;
}
return ans;
}
}

• class Solution {
public:
vector<int> getMaximumXor(vector<int>& nums, int maximumBit) {
int xs = 0;
for (int& x : nums) {
xs ^= x;
}
int n = nums.size();
vector<int> ans(n);
for (int i = 0; i < n; ++i) {
int x = nums[n - i - 1];
int k = 0;
for (int j = maximumBit - 1; ~j; --j) {
if ((xs >> j & 1) == 0) {
k |= 1 << j;
}
}
ans[i] = k;
xs ^= x;
}
return ans;
}
};

• class Solution:
def getMaximumXor(self, nums: List[int], maximumBit: int) -> List[int]:
ans = []
xs = reduce(xor, nums)
for x in nums[::-1]:
k = 0
for i in range(maximumBit - 1, -1, -1):
if (xs >> i & 1) == 0:
k |= 1 << i
ans.append(k)
xs ^= x
return ans


• func getMaximumXor(nums []int, maximumBit int) (ans []int) {
xs := 0
for _, x := range nums {
xs ^= x
}
for i := range nums {
x := nums[len(nums)-i-1]
k := 0
for j := maximumBit - 1; j >= 0; j-- {
if xs>>j&1 == 0 {
k |= 1 << j
}
}
ans = append(ans, k)
xs ^= x
}
return
}

• function getMaximumXor(nums: number[], maximumBit: number): number[] {
let xs = 0;
for (const x of nums) {
xs ^= x;
}
const n = nums.length;
const ans = new Array(n);
for (let i = 0; i < n; ++i) {
const x = nums[n - i - 1];
let k = 0;
for (let j = maximumBit - 1; j >= 0; --j) {
if (((xs >> j) & 1) == 0) {
k |= 1 << j;
}
}
ans[i] = k;
xs ^= x;
}
return ans;
}


• /**
* @param {number[]} nums
* @param {number} maximumBit
* @return {number[]}
*/
var getMaximumXor = function (nums, maximumBit) {
let xs = 0;
for (const x of nums) {
xs ^= x;
}
const n = nums.length;
const ans = new Array(n);
for (let i = 0; i < n; ++i) {
const x = nums[n - i - 1];
let k = 0;
for (let j = maximumBit - 1; j >= 0; --j) {
if (((xs >> j) & 1) == 0) {
k |= 1 << j;
}
}
ans[i] = k;
xs ^= x;
}
return ans;
};


• public class Solution {
public int[] GetMaximumXor(int[] nums, int maximumBit) {
int xs = 0;
foreach (int x in nums) {
xs ^= x;
}
int n = nums.Length;
int[] ans = new int[n];
for (int i = 0; i < n; ++i) {
int x = nums[n - i - 1];
int k = 0;
for (int j = maximumBit - 1; j >= 0; --j) {
if ((xs >> j & 1) == 0) {
k |= 1 << j;
}
}
ans[i] = k;
xs ^= x;
}
return ans;
}
}