Maximum XOR for Each Query

class Solution {
public:
    vector<int> calculateMaximumXor(vector<int>& values, int bitLevel) {
        int currentXor = 0;
        for (int num : values) {
            currentXor ^= num;
        }
        vector<int> result(values.size());
    
        int fullMask = (1 << bitLevel) - 1;
    
        for (int index = 0; index < values.size(); index++) {
            result[index] = currentXor ^ fullMask;
            // Reverse through list reducing xor
            currentXor ^= values[values.size() - 1 - index];
        }
    
        return result;
    }
};

This solution describes a method to determine the maximum XOR for each query from a list of integers and a specified bit level using C++. The approach leverages bit manipulation to achieve the results.

Firstly, the function starts by calculating the XOR of all elements in the provided list values. It stores this result in currentXor.

Next, it calculates the full bit mask for the given bitLevel. This mask is computed as (1 << bitLevel) - 1, which effectively sets all bits less than bitLevel to 1.

The function then iterates over each element in values. For each index, it computes the XOR of currentXor with fullMask and stores this in the result list. Essentially, this operation aims to flip the bits of currentXor up to the specified bit level.

To update currentXor for the next iteration continuously, the algorithm reverses through the values list while reducing the XOR operation by the current values from the end of the list. This reverse XOR operation guarantees that each query's resultant XOR reflects the loss of the specific integer's bit influence from the calculations.

Finally, the function returns the results as a list of integers where each integer represents the maximum XOR possible with the remaining integers in the list by the end of each query. This solution efficiently utilizes bitwise operations for optimal performance in scenarios involving bit manipulations in queries.

class Solution {
    
    public int[] computeMaxXor(int[] values, int maxBit) {
        int cumulativeXor = 0;
        for (int v : values) {
            cumulativeXor ^= v;
        }
        int[] result = new int[values.length];
    
        int fullMask = (1 << maxBit) - 1;
    
        for (int idx = 0; idx < values.length; idx++) {
            result[idx] = cumulativeXor ^ fullMask;
            cumulativeXor ^= values[values.length - 1 - idx];
        }
    
        return result;
    }
}

The Java code provided outlines a solution for computing the maximum XOR for each query based on a given array values and an integer maxBit.

• The method computeMaxXor accepts two parameters:

  • int[] values - an array of integers.
  • int maxBit - the maximum bit position to consider in the XOR calculations.

• The code uses an integer cumulativeXor initialized to 0, to accumulate the XOR of all elements in the values array, ensuring that all bits up to maxBit are considered.

• An integer array result with the same length as values is initialized to store the result of each XOR calculation against the full mask.

• int fullMask is calculated using 1 << maxBit, subtracted by 1. This operation creates a binary number that consists entirely of 1's up to the maxBitth position.

• A loop runs through the values array:

  • At each index idx, the XOR of cumulativeXor with fullMask is stored in result[idx].
  • Then, cumulativeXor is updated by XORing it with the element in values starting from the end towards the beginning (values[values.length - 1 - idx]).

• The method finally returns the result array, which contains the maximum XOR results for each query, considering the sequence and the constraints set by maxBit.

This approach efficiently computes the required values by progressively accumulating the XOR and applying it against a full mask of significant bits. The use of bitwise operations ensures the method is optimized for quick computation.

class Solution:
    def calculateMaxXor(self, num_list: List[int], max_bits: int) -> List[int]:
        cumulative_xor = 0
        for num in num_list:
            cumulative_xor ^= num
        result = [0] * len(num_list)
    
        all_ones_mask = (1 << max_bits) - 1
    
        for index in range(len(num_list)):
            result[index] = cumulative_xor ^ all_ones_mask
            cumulative_xor ^= num_list[len(num_list) - 1 - index]
    
        return result

The given Python code defines a method to compute the maximum XOR for each query from a list of integers using a bitwise operation approach. The function calculateMaxXor in the Solution class takes two parameters: num_list, a list of integers, and max_bits, which represents the number of bits to consider for the maximum possible value.

• Initialize cumulative_xor to zero. This variable will store the cumulative XOR of all numbers processed so far.
• Based on max_bits, compute all_ones_mask which is a number consisting of all 1s of length equal to max_bits.
• Iterate through each number in the num_list:
  • Perform XOR between cumulative_xor and all_ones_mask to store in the result list for the current index.
  • Update cumulative_xor by XORing it with the current number from the end of the list moving backward.
• Finally, return the result list containing maximum XOR values after each query.

This algorithm effectively utilizes cumulative XOR values and a mask of all ones to find the desired result and adjusts the cumulative XOR in reverse to cater for each individual query dynamically. This method ensures that each XOR result is calculated using an efficient bitwise manipulation, achieving the task in an optimized manner.