LeetCode #1829 — MEDIUM

Maximum XOR for Each Query

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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:

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

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

Example 1:

Input: nums = [0,1,1,3], maximumBit = 2
Output: [0,3,2,3]
Explanation: The queries are answered as follows:
1^st query: nums = [0,1,1,3], k = 0 since 0 XOR 1 XOR 1 XOR 3 XOR 0 = 3.
2^nd query: nums = [0,1,1], k = 3 since 0 XOR 1 XOR 1 XOR 3 = 3.
3^rd query: nums = [0,1], k = 2 since 0 XOR 1 XOR 2 = 3.
4^th 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:
1^st query: nums = [2,3,4,7], k = 5 since 2 XOR 3 XOR 4 XOR 7 XOR 5 = 7.
2^nd query: nums = [2,3,4], k = 2 since 2 XOR 3 XOR 4 XOR 2 = 7.
3^rd query: nums = [2,3], k = 6 since 2 XOR 3 XOR 6 = 7.
4^th 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] < 2^maximumBit
nums is sorted in ascending order.

Step 01

Brute Force Baseline

Problem summary: 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: 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. Remove the last element from the current array nums. Return an array answer, where answer[i] is the answer to the ith query.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Bit Manipulation

Example 1

[0,1,1,3]
2

Example 2

[2,3,4,7]
3

Example 3

[0,1,2,2,5,7]
3

Core Insight

What unlocks the optimal approach

Note that the maximum possible XOR result is always 2^(maximumBit) - 1
So the answer for a prefix is the XOR of that prefix XORed with 2^(maximumBit)-1

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1829: Maximum XOR for Each Query
class Solution {
    public int[] getMaximumXor(int[] nums, int maximumBit) {
        int n = nums.length;
        int xs = 0;
        for (int x : nums) {
            xs ^= x;
        }
        int[] ans = new int[n];
        for (int i = 0; i < n; ++i) {
            int x = nums[n - i - 1];
            int k = 0;
            for (int j = maximumBit - 1; j >= 0; --j) {
                if (((xs >> j) & 1) == 0) {
                    k |= 1 << j;
                }
            }
            ans[i] = k;
            xs ^= x;
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1829: Maximum XOR for Each Query
func getMaximumXor(nums []int, maximumBit int) (ans []int) {
	xs := 0
	for _, x := range nums {
		xs ^= x
	}
	for i := range nums {
		x := nums[len(nums)-i-1]
		k := 0
		for j := maximumBit - 1; j >= 0; j-- {
			if xs>>j&1 == 0 {
				k |= 1 << j
			}
		}
		ans = append(ans, k)
		xs ^= x
	}
	return
}

# Accepted solution for LeetCode #1829: Maximum XOR for Each Query
class Solution:
    def getMaximumXor(self, nums: List[int], maximumBit: int) -> List[int]:
        ans = []
        xs = reduce(xor, nums)
        for x in nums[::-1]:
            k = 0
            for i in range(maximumBit - 1, -1, -1):
                if (xs >> i & 1) == 0:
                    k |= 1 << i
            ans.append(k)
            xs ^= x
        return ans

// Accepted solution for LeetCode #1829: Maximum XOR for Each Query
struct Solution;

impl Solution {
    fn get_maximum_xor(nums: Vec<i32>, maximum_bit: i32) -> Vec<i32> {
        let mask: i32 = (1 << maximum_bit) - 1;
        let mut nums: Vec<i32> = nums.into_iter().map(|x| x & mask).collect();
        let mut xor = nums.iter().fold(0, |acc, x| acc ^ x);
        let mut res = vec![];
        while let Some(last) = nums.pop() {
            res.push(mask ^ xor);
            xor ^= last;
        }
        res
    }
}

#[test]
fn test() {
    let nums = vec![0, 1, 1, 3];
    let maximum_bit = 2;
    let res = vec![0, 3, 2, 3];
    assert_eq!(Solution::get_maximum_xor(nums, maximum_bit), res);
    let nums = vec![2, 3, 4, 7];
    let maximum_bit = 3;
    let res = vec![5, 2, 6, 5];
    assert_eq!(Solution::get_maximum_xor(nums, maximum_bit), res);
    let nums = vec![0, 1, 2, 2, 5, 7];
    let maximum_bit = 3;
    let res = vec![4, 3, 6, 4, 6, 7];
    assert_eq!(Solution::get_maximum_xor(nums, maximum_bit), res);
}

// Accepted solution for LeetCode #1829: Maximum XOR for Each Query
function getMaximumXor(nums: number[], maximumBit: number): number[] {
    let xs = 0;
    for (const x of nums) {
        xs ^= x;
    }
    const n = nums.length;
    const ans = new Array(n);
    for (let i = 0; i < n; ++i) {
        const x = nums[n - i - 1];
        let k = 0;
        for (let j = maximumBit - 1; j >= 0; --j) {
            if (((xs >> j) & 1) == 0) {
                k |= 1 << j;
            }
        }
        ans[i] = k;
        xs ^= x;
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

SORT + SCAN

O(n log n) time

O(n) space

Sort the array in O(n log n), then scan for the missing or unique element by comparing adjacent pairs. Sorting requires O(n) auxiliary space (or O(1) with in-place sort but O(n log n) time remains). The sort step dominates.

BIT MANIPULATION

O(n) time

O(1) space

Bitwise operations (AND, OR, XOR, shifts) are O(1) per operation on fixed-width integers. A single pass through the input with bit operations gives O(n) time. The key insight: XOR of a number with itself is 0, which eliminates duplicates without extra space.

Shortcut: Bit operations are O(1). XOR cancels duplicates. Single pass → O(n) time, O(1) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.