LeetCode 1829: Maximum XOR for Each Query Solution

Master LeetCode problem 1829 (Maximum XOR for Each Query), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1829. Maximum XOR for Each Query

Medium

Array Bit Manipulation Prefix Sum

Problem Explanation

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:

Solution Code

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;
    }
}

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