Maximum XOR for Each Query

Problem 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[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.

Key Insights

• The goal is to maximize the XOR of the entire array with an additional number k.
• The maximum possible value for k can be calculated as 2^maximumBit - 1, which helps in determining the optimal k.
• The XOR operation has properties such that for each bit position, we can decide whether to set that bit in k based on the cumulative XOR of the elements in nums.

Space and Time Complexity

Time Complexity: O(n)
Space Complexity: O(1)

Solution

To solve the problem, we use the following approach:

1. Calculate the cumulative XOR of all elements in the nums array. This gives us a base value that we'll use to determine the optimal k for maximizing the XOR.
2. Determine the maximum possible value of k as maxK = (1 << maximumBit) - 1.
3. For each query, compute k as maxK XOR cumulativeXOR. This is because maximizing the result means we want to flip bits in the cumulative XOR to reach maxK.
4. Store the result and then remove the last element of the array for the next iteration.

This process is repeated until all elements are processed.

Code Solutions