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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]

 

Constraints:

  • nums.length == n
  • 1 <= n <= 105
  • 1 <= maximumBit <= 20
  • 0 <= nums[i] < 2maximumBit
  • nums​​​ is sorted in ascending order.

Solution Explanation for LeetCode 1829: Maximum XOR for Each Query

This problem asks us to find the maximum XOR value achievable by XORing the current array with a number k, and then remove the last element, repeating this process.

Core Idea: The key to efficiently solving this problem lies in understanding XOR properties. The XOR operation is commutative and associative. Furthermore, x ^ x == 0 for any x. This means that if we have the XOR sum of the entire array, removing an element is simply XORing that element back into the sum.

Algorithm (Solution 2 - Optimized):

  1. Calculate Initial XOR Sum: First, we calculate the XOR sum of all elements in nums. Let's call this xs.

  2. Create Mask: We determine the maximum possible value of k which is (1 << maximumBit) - 1. This is a bitmask with all bits set to 1 up to maximumBit. We will call this mask.

  3. Iterate and XOR: We iterate through the array nums from right to left. For each element x:

    • The maximum XOR achievable is obtained by XORing the current xs with the mask. This flips all bits in xs resulting in the maximum value given the constraint. This is k = xs ^ mask.
    • We append k to our result list ans.
    • We update xs by XORing the current element x: xs ^= x. This effectively removes x from the XOR sum.

  4. Return Result: Finally, we return the ans list, containing the k values for each query.

Time Complexity Analysis:

  • The initial XOR sum calculation takes O(n) time.
  • The iteration and XOR operations also take O(n) time.
  • Therefore, the overall time complexity is O(n), linear with respect to the input array size.

Space Complexity Analysis:

  • The space used is dominated by the ans list, which stores n integers.
  • Therefore, the space complexity is O(n).

Code (Python - Solution 2):

def getMaximumXor(nums, maximumBit):
    """
    Efficiently finds the maximum XOR for each query.
 
    Args:
        nums: A sorted list of non-negative integers.
        maximumBit: The maximum number of bits allowed for k.
 
    Returns:
        A list of integers representing the answers to each query.
    """
    xs = 0  # Initialize XOR sum
    for x in nums:
        xs ^= x
 
    mask = (1 << maximumBit) - 1  # Create bitmask
    ans = []
    for x in reversed(nums): # Iterate from right to left
        k = xs ^ mask         # Calculate k for maximum XOR
        ans.append(k)
        xs ^= x                # Remove x from XOR sum
 
    return ans

This optimized solution avoids bit-by-bit manipulation, significantly improving efficiency. The other solution (Solution 1) is provided in the original response, demonstrating a less efficient, bit-by-bit approach. The time complexity of Solution 1 is O(n*m) where m is maximumBit. Solution 2 is preferred for its better time complexity.