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1829. Maximum XOR for Each Query
Description
You are given a sorted array nums
of n
nonnegative integers and an integer maximumBit
. You want to perform the following query n
times:
 Find a nonnegative integer
k < 2^{maximumBit}
such thatnums[0] XOR nums[1] XOR ... XOR nums[nums.length1] XOR k
is maximized.k
is the answer to thei^{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: 1^{st} query: nums = [0,1,1,3], k = 0 since 0 XOR 1 XOR 1 XOR 3 XOR 0 = 3. 2^{nd} query: nums = [0,1,1], k = 3 since 0 XOR 1 XOR 1 XOR 3 = 3. 3^{rd} query: nums = [0,1], k = 2 since 0 XOR 1 XOR 2 = 3. 4^{th} 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: 1^{st} query: nums = [2,3,4,7], k = 5 since 2 XOR 3 XOR 4 XOR 7 XOR 5 = 7. 2^{nd} query: nums = [2,3,4], k = 2 since 2 XOR 3 XOR 4 XOR 2 = 7. 3^{rd} query: nums = [2,3], k = 6 since 2 XOR 3 XOR 6 = 7. 4^{th} 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.
Solutions
Solution 1: Bitwise Operation + Enumeration
First, we preprocess the XOR sum $xs$ of the array nums
, i.e., $xs=nums[0] \oplus nums[1] \oplus \cdots \oplus nums[n1]$.
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., $xs=nums[0] \oplus nums[1] \oplus \cdots \oplus nums[n1]$.
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)i1] 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; } }