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

# 1829. Maximum XOR for Each Query

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:

1. Find a non-negative integer k < 2^maximumBit such that nums XOR nums XOR ... XOR nums[nums.length-1] XOR k is maximized. k is the answer to the i-th query.
2. 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 = , 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 = , 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 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 = nums;
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;
}
}

############

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

• // OJ: https://leetcode.com/problems/maximum-xor-for-each-query/
// Time: O(N)
// Space: O(1)
class Solution {
public:
vector<int> getMaximumXor(vector<int>& A, int MB) {
int N = A.size(), t = (1 << MB) - 1;
vector<int> ans(N);
for (int i = 0; i < N; ++i) {
A[i] ^= i == 0 ? 0 : A[i - 1];
ans[N - i - 1] = t ^ A[i];
}
return ans;
}
};

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

############

# 1829. Maximum XOR for Each Query
# https://leetcode.com/problems/maximum-xor-for-each-query/

class Solution:
def getMaximumXor(self, nums: List[int], maximumBit: int) -> List[int]:
n = len(nums)
m = (2 ** maximumBit) - 1
res = []

for i in range(1, n):
nums[i] ^= nums[i - 1]

for num in nums[::-1]:
res.append(num ^ m)

return res


• func getMaximumXor(nums []int, maximumBit int) (ans []int) {
xs := 0
for _, x := range nums {
xs ^= x
}
mask := (1 << maximumBit) - 1
for i := range nums {
x := nums[len(nums)-i-1]
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 mask = (1 << maximumBit) - 1;
const n = nums.length;
const ans = new Array(n);
for (let i = 0; i < n; ++i) {
const x = nums[n - i - 1];
let k = xs ^ mask;
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 mask = (1 << maximumBit) - 1;
const n = nums.length;
const ans = new Array(n);
for (let i = 0; i < n; ++i) {
const x = nums[n - i - 1];
let k = xs ^ mask;
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 mask = (1 << maximumBit) - 1;
int n = nums.Length;
int[] ans = new int[n];
for (int i = 0; i < n; ++i) {
int x = nums[n - i - 1];
int k = xs ^ mask;
ans[i] = k;
xs ^= x;
}
return ans;
}
}