LeetCode 1829: Maximum XOR for Each Query

Explanation:

• We need to find a number k such that XOR of all elements in the given array along with k is maximized.
• To maximize XOR, we need to consider the bits at each position in a binary representation.
• Since the array is sorted, we can iterate from the highest bit to the lowest bit.
• For each bit position, we check how many set bits are there in the current prefix XOR of the array. We can then set the bit at that position in k to 1 if it helps in maximizing XOR.
• Finally, we return the calculated k for each query. Solution:

class Solution {
    public int[] getMaximumXor(int[] nums, int maximumBit) {
        int n = nums.length;
        int totalXOR = 0;
        for (int num : nums) {
            totalXOR ^= num;
        }
        
        int[] answer = new int[n];
        int mask = (1 << maximumBit) - 1;
        
        for (int i = n - 1; i >= 0; i--) {
            answer[n - 1 - i] = totalXOR ^ mask;
            totalXOR ^= nums[i];
        }
        
        return answer;
    }
}