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.
1function maxXorQueries(nums, maximumBit) {
2 const maxNum = (1 << maximumBit) - 1;
3 let totalXor = 0;
4 const result = new Array(nums.length);
5 for (const num of nums) {
6 totalXor ^= num;
7 }
8 for (let i = 0; i < nums.length; ++i) {
9 result[i] = totalXor ^ maxNum;
10 totalXor ^= nums[nums.length - 1 - i];
11 }
12 return result;
13}
14
15const nums = [0, 1, 1, 3];
16const maximumBit = 2;
17console.log(maxXorQueries(nums, maximumBit));This function follows the same computation strategy: tracking the XOR of all numbers, adjusting it as each number is evaluated in decreasing index order for each query.
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.
1def maxXorQueriesPrefix(nums, maximumBit):
2 max_num = (1 << maximumBit) - 1
3 prefix_xor = [0] * len(nums)
4 result = []
5 prefix_xor[0] = nums[0]
6 for i in range(1, len(nums)):
7 prefix_xor[i] = prefix_xor[i - 1] ^ nums[i]
8 for i in range(len(nums)):
9 current_xor = prefix_xor[-1] if i == 0 else prefix_xor[-1] ^ prefix_xor[i - 1]
10 result.append(current_xor ^ max_num)
11 return result
12
13nums = [0, 1, 1, 3]
14maximumBit = 2
15print(maxXorQueriesPrefix(nums, maximumBit))Here, maxXorQueriesPrefix precomputes a prefix XOR list. The computation and determination of maximum XOR value is done efficiently due to cumulative XOR properties enabled by using prefix XOR.