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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 $O(1)$.

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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

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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;
    }
}

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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;
    }
};

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func getMaximumXor(nums []int, maximumBit int) (ans []int) {
    xs := 0
    for _, x