In this approach, we compute the total XOR of the array and then, for each query, determine the best integer k to maximize the XOR. We leverage the fact that XOR of a number with itself is 0, and XOR with 0 is the number itself. The trick is to use the maximum number in the range [0, 2^maximumBit) as an XOR mask to get the maximum number at each step.
We'll precompute the total XOR of the given array. For each step, find k by XORing the current XOR value with the maximum possible number, which is `(2^maximumBit - 1)`. After each query, remove the last element and update the XOR till it becomes empty.
Time Complexity: O(n), where n is the length of the array.
Space Complexity: O(1) additional space beyond input and output arrays.
1using System;
2
3class MaxXorQueries {
4 public static int[] MaxXorQueries(int[] nums, int maximumBit) {
5 int maxNum = (1 << maximumBit) - 1;
6 int totalXor = 0;
7 int[] result = new int[nums.Length];
8 foreach (int num in nums) {
9 totalXor ^= num;
10 }
11 for (int i = 0; i < nums.Length; ++i) {
12 result[i] = totalXor ^ maxNum;
13 totalXor ^= nums[nums.Length - 1 - i];
14 }
15 return result;
16 }
17
18 public static void Main(string[] args) {
19 int[] nums = {0, 1, 1, 3};
20 int maximumBit = 2;
21 int[] result = MaxXorQueries(nums, maximumBit);
22 Console.WriteLine(string.Join(" ", result));
23 }
24}The method MaxXorQueries implements XOR operations for the entire list and then removes each number's influence as queries proceed, maintaining a result list to store outcomes for each question asked.
This approach calculates the XOR using prefix sums for XOR which allows direct computation of the desired XOR values for any subset by removing the influence of past elements efficiently. We combine this with the known max number XOR mask strategy to derive each query's answer.
First, we compute the prefix XOR array which gives cumulative XOR up to any index. By doing so, calculating new ranges becomes direct through array subtraction for the prefix array indices, because of the property (A XOR A = 0) property that cancels out all elements between boundaries.
Time Complexity: O(n)
Space Complexity: O(n) for the prefix array.
1#include <stdio.h>
2
3void maxXorQueriesPrefix(int *nums, int n, int maximumBit, int *result) {
4 int maxNum = (1 << maximumBit) - 1;
5 int prefixXor[n];
6 prefixXor[0] = nums[0];
7 for (int i = 1; i < n; ++i) {
8 prefixXor[i] = prefixXor[i - 1] ^ nums[i];
9 }
10 for (int i = 0; i < n; ++i) {
11 int currentXor = (i == 0) ? prefixXor[n - 1] : prefixXor[n - 1] ^ prefixXor[i - 1];
12 result[i] = currentXor ^ maxNum;
13 }
14}
15
16int main() {
17 int nums[] = {0, 1, 1, 3};
18 int maximumBit = 2;
19 int n = sizeof(nums) / sizeof(nums[0]);
20 int result[n];
21 maxXorQueriesPrefix(nums, n, maximumBit, result);
22 for (int i = 0; i < n; i++) {
23 printf("%d ", result[i]);
24 }
25 return 0;
26}This code uses a prefix XOR array that keeps track of all XOR values up to each index. The result for each query, which requests the maximum XOR when combinations incorporate fewer elements, is calculated based on these cumulative XOR values adjusted with the maximum potential XOR mask before appending results.