Maximum XOR for Each Query

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]

Maximizing XOR Queries in an Array

This problem challenges us to find a k value that maximizes the XOR sum of an array with k, and then iteratively remove elements and repeat the process. Let's explore a solution.

1. Naive Approach (Brute Force)

A brute-force approach would involve iterating through all possible values of k (from 0 to 2^maximumBit - 1) for each query and calculating the XOR sum to find the maximum. This is inefficient, especially for larger arrays and maximumBit values.

def solve_naive(nums, maximumBit):
    answer = []
    while nums:
        xor_sum = 0
        for num in nums:
            xor_sum ^= num
        
        max_xor = -1
        best_k = -1
        for k in range(2**maximumBit):
            current_xor = xor_sum ^ k
            if current_xor > max_xor:
                max_xor = current_xor
                best_k = k
        
        answer.append(best_k)
        nums.pop()
    return answer

2. Optimal Approach

We can optimize this by pre-calculating the XOR sum of the array. For each query, we can find the k that maximizes the XOR sum by XORing the current XOR sum with the maximum possible value (2^maximumBit - 1).

def solve_optimal(nums, maximumBit):
    answer = []
    xor_sum = 0
    for num in nums:
        xor_sum ^= num
    
    max_val = (1 << maximumBit) - 1  # Maximum possible value

    for _ in range(len(nums)):
        k = max_val ^ xor_sum
        answer.append(k)
        xor_sum ^= nums.pop()
    return answer

3. Example

Let's trace the solve_optimal function with nums = [0, 1, 1, 3] and maximumBit = 2:

1. Initialization: xor_sum = 0
2. Calculate initial XOR sum:
   • xor_sum ^= 0 => xor_sum = 0
   • xor_sum ^= 1 => xor_sum = 1
   • xor_sum ^= 1 => xor_sum = 0
   • xor_sum ^= 3 => xor_sum = 3
3. max_val = (1 << 2) - 1 = 3
4. Loop:
   • k = 3 ^ 3 = 0, answer.append(0), xor_sum ^= 3 => xor_sum = 0, nums = [0, 1, 1]
   • k = 3 ^ 0 = 3, answer.append(3), xor_sum ^= 1 => xor_sum = 1, nums = [0, 1]
   • k = 3 ^ 1 = 2, answer.append(2), xor_sum ^= 1 => xor_sum = 0, nums = [0]
   • k = 3 ^ 0 = 3, answer.append(3), xor_sum ^= 0 => xor_sum = 0, nums = []
5. Return: [0, 3, 2, 3]

4. Big(O) Time Complexity Analysis

The optimal solution has a time complexity of O(n), where n is the length of the input array nums. This is because we iterate through the array once to calculate the initial XOR sum and then iterate again to find the k values and remove elements. Each operation within the loops takes constant time.

5. Big(O) Space Complexity Analysis

The space complexity of the optimal solution is O(1) because we only use a few extra variables (xor_sum, max_val, k) that do not depend on the size of the input array. The answer list is technically O(n), but is required by the output, and doesn't count towards extra space complexity.

6. Edge Cases

• Empty Array: If the input array nums is empty, the algorithm should return an empty array as the answer.
• maximumBit is 0: If maximumBit is 0, then the only possible value for k is 0. The algorithm should handle this case correctly.
• Large maximumBit: The algorithm should work correctly for large values of maximumBit as long as the intermediate calculations do not overflow.

7. Complete Code

def solve(nums, maximumBit):
    answer = []
    xor_sum = 0
    for num in nums:
        xor_sum ^= num
    
    max_val = (1 << maximumBit) - 1  # Maximum possible value

    for _ in range(len(nums)):
        k = max_val ^ xor_sum
        answer.append(k)
        xor_sum ^= nums.pop()
    return answer

# Example Usage
nums = [0, 1, 1, 3]
maximumBit = 2
result = solve(nums, maximumBit)
print(result)  # Output: [0, 3, 2, 3]