README.md

1829. Maximum XOR for Each Query

You are given a sorted array nums of n non-negative integers and an integer maximumBit. You want to perform the following query n times:

1. Find a non-negative integer k < 2maximumBit such that nums[0] XOR nums[1] XOR ... XOR nums[nums.length-1] XOR k is maximized. k is the answer to the ith query.
2. Remove the last element from the current array nums.

Return an array answer, where answer[i] is the answer to the ith query.

Example 1:

Input: nums = [0,1,1,3], maximumBit = 2
Output: [0,3,2,3]
Explanation: The queries are answered as follows:
1st query: nums = [0,1,1,3], k = 0 since 0 XOR 1 XOR 1 XOR 3 XOR 0 = 3.
2nd query: nums = [0,1,1], k = 3 since 0 XOR 1 XOR 1 XOR 3 = 3.
3rd query: nums = [0,1], k = 2 since 0 XOR 1 XOR 2 = 3.
4th query: nums = [0], k = 3 since 0 XOR 3 = 3.

Example 2:

Input: nums = [2,3,4,7], maximumBit = 3
Output: [5,2,6,5]
Explanation: The queries are answered as follows:
1st query: nums = [2,3,4,7], k = 5 since 2 XOR 3 XOR 4 XOR 7 XOR 5 = 7.
2nd query: nums = [2,3,4], k = 2 since 2 XOR 3 XOR 4 XOR 2 = 7.
3rd query: nums = [2,3], k = 6 since 2 XOR 3 XOR 6 = 7.
4th query: nums = [2], k = 5 since 2 XOR 5 = 7.

Example 3:

Input: nums = [0,1,2,2,5,7], maximumBit = 3
Output: [4,3,6,4,6,7]

Constraints:

• nums.length == n
• 1 <= n <= 105
• 1 <= maximumBit <= 20
• 0 <= nums[i] < 2maximumBit
• nums is sorted in ascending order.

Solutions (Rust)

1. Solution

impl Solution {
    pub fn get_maximum_xor(nums: Vec<i32>, maximum_bit: i32) -> Vec<i32> {
        let mut xor = nums.iter().fold(0, |acc, x| acc ^ x);
        let mut answer = vec![(1 << maximum_bit) - 1; nums.len()];

        for i in 0..nums.len() {
            answer[i] ^= xor;
            xor ^= nums[nums.len() - 1 - i];
        }

        answer
    }
}