You are given a sorted array nums of n non-negative integers and an integer maximumBit. You want to perform the following query ntimes:
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.
Remove the last element from the current array nums.
Return an arrayanswer, 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.
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 $O(1)$.
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 $O(1)$.