LeetCode #1707 — HARD

Maximum XOR With an Element From Array

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

You are given an array nums consisting of non-negative integers. You are also given a queries array, where queries[i] = [x_i, m_i].

The answer to the i^th query is the maximum bitwise XOR value of x_i and any element of nums that does not exceed m_i. In other words, the answer is max(nums[j] XOR x_i) for all j such that nums[j] <= m_i. If all elements in nums are larger than m_i, then the answer is -1.

Return an integer array answer where answer.length == queries.length and answer[i] is the answer to the i^th query.

Example 1:

Input: nums = [0,1,2,3,4], queries = [[3,1],[1,3],[5,6]]
Output: [3,3,7]
Explanation:
1) 0 and 1 are the only two integers not greater than 1. 0 XOR 3 = 3 and 1 XOR 3 = 2. The larger of the two is 3.
2) 1 XOR 2 = 3.
3) 5 XOR 2 = 7.

Example 2:

Input: nums = [5,2,4,6,6,3], queries = [[12,4],[8,1],[6,3]]
Output: [15,-1,5]

Constraints:

1 <= nums.length, queries.length <= 10⁵
queries[i].length == 2
0 <= nums[j], x_i, m_i <= 10⁹

Step 01

Brute Force Baseline

Problem summary: You are given an array nums consisting of non-negative integers. You are also given a queries array, where queries[i] = [xi, mi]. The answer to the ith query is the maximum bitwise XOR value of xi and any element of nums that does not exceed mi. In other words, the answer is max(nums[j] XOR xi) for all j such that nums[j] <= mi. If all elements in nums are larger than mi, then the answer is -1. Return an integer array answer where answer.length == queries.length and 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 · Trie

Example 1

[0,1,2,3,4]
[[3,1],[1,3],[5,6]]

Example 2

[5,2,4,6,6,3]
[[12,4],[8,1],[6,3]]

Core Insight

What unlocks the optimal approach

In problems involving bitwise operations, we often think on the bits level. In this problem, we can think that to maximize the result of an xor operation, we need to maximize the most significant bit, then the next one, and so on.
If there's some number in the array that is less than m and whose the most significant bit is different than that of x, then xoring with this number maximizes the most significant bit, so I know this bit in the answer is 1.
To check the existence of such numbers and narrow your scope for further bits based on your choice, you can use trie.
You can sort the array and the queries, and maintain the trie such that in each query the trie consists exactly of the valid elements.

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

Largest constraint values

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 #1707: Maximum XOR With an Element From Array
class Trie {
    private Trie[] children = new Trie[2];

    public void insert(int x) {
        Trie node = this;
        for (int i = 30; i >= 0; --i) {
            int v = x >> i & 1;
            if (node.children[v] == null) {
                node.children[v] = new Trie();
            }
            node = node.children[v];
        }
    }

    public int search(int x) {
        Trie node = this;
        int ans = 0;
        for (int i = 30; i >= 0; --i) {
            int v = x >> i & 1;
            if (node.children[v ^ 1] != null) {
                ans |= 1 << i;
                node = node.children[v ^ 1];
            } else if (node.children[v] != null) {
                node = node.children[v];
            } else {
                return -1;
            }
        }
        return ans;
    }
}

class Solution {
    public int[] maximizeXor(int[] nums, int[][] queries) {
        Arrays.sort(nums);
        int n = queries.length;
        Integer[] idx = new Integer[n];
        Arrays.setAll(idx, i -> i);
        Arrays.sort(idx, (i, j) -> queries[i][1] - queries[j][1]);
        int[] ans = new int[n];
        Trie trie = new Trie();
        int j = 0;
        for (int i : idx) {
            int x = queries[i][0], m = queries[i][1];
            while (j < nums.length && nums[j] <= m) {
                trie.insert(nums[j++]);
            }
            ans[i] = trie.search(x);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1707: Maximum XOR With an Element From Array
type Trie struct {
	children [2]*Trie
}

func NewTrie() *Trie {
	return &Trie{}
}

func (t *Trie) insert(x int) {
	node := t
	for i := 30; i >= 0; i-- {
		v := x >> i & 1
		if node.children[v] == nil {
			node.children[v] = NewTrie()
		}
		node = node.children[v]
	}
}

func (t *Trie) search(x int) int {
	node := t
	ans := 0
	for i := 30; i >= 0; i-- {
		v := x >> i & 1
		if node.children[v^1] != nil {
			ans |= 1 << i
			node = node.children[v^1]
		} else if node.children[v] != nil {
			node = node.children[v]
		} else {
			return -1
		}
	}
	return ans
}

func maximizeXor(nums []int, queries [][]int) []int {
	sort.Ints(nums)
	n := len(queries)
	idx := make([]int, n)
	for i := 0; i < n; i++ {
		idx[i] = i
	}
	sort.Slice(idx, func(i, j int) bool {
		return queries[idx[i]][1] < queries[idx[j]][1]
	})
	ans := make([]int, n)
	trie := NewTrie()
	j := 0
	for _, i := range idx {
		x, m := queries[i][0], queries[i][1]
		for j < len(nums) && nums[j] <= m {
			trie.insert(nums[j])
			j++
		}
		ans[i] = trie.search(x)
	}
	return ans
}

# Accepted solution for LeetCode #1707: Maximum XOR With an Element From Array
class Trie:
    __slots__ = ["children"]

    def __init__(self):
        self.children = [None] * 2

    def insert(self, x: int):
        node = self
        for i in range(30, -1, -1):
            v = x >> i & 1
            if node.children[v] is None:
                node.children[v] = Trie()
            node = node.children[v]

    def search(self, x: int) -> int:
        node = self
        ans = 0
        for i in range(30, -1, -1):
            v = x >> i & 1
            if node.children[v ^ 1]:
                ans |= 1 << i
                node = node.children[v ^ 1]
            elif node.children[v]:
                node = node.children[v]
            else:
                return -1
        return ans


class Solution:
    def maximizeXor(self, nums: List[int], queries: List[List[int]]) -> List[int]:
        trie = Trie()
        nums.sort()
        j, n = 0, len(queries)
        ans = [-1] * n
        for i, (x, m) in sorted(zip(range(n), queries), key=lambda x: x[1][1]):
            while j < len(nums) and nums[j] <= m:
                trie.insert(nums[j])
                j += 1
            ans[i] = trie.search(x)
        return ans

// Accepted solution for LeetCode #1707: Maximum XOR With an Element From Array
/**
 * [1707] Maximum XOR With an Element From Array
 *
 * You are given an array nums consisting of non-negative integers. You are also given a queries array, where queries[i] = [xi, mi].
 * The answer to the i^th query is the maximum bitwise XOR value of xi and any element of nums that does not exceed mi. In other words, the answer is max(nums[j] XOR xi) for all j such that nums[j] <= mi. If all elements in nums are larger than mi, then the answer is -1.
 * Return an integer array answer where answer.length == queries.length and answer[i] is the answer to the i^th query.
 *  
 * Example 1:
 *
 * Input: nums = [0,1,2,3,4], queries = [[3,1],[1,3],[5,6]]
 * Output: [3,3,7]
 * Explanation:
 * 1) 0 and 1 are the only two integers not greater than 1. 0 XOR 3 = 3 and 1 XOR 3 = 2. The larger of the two is 3.
 * 2) 1 XOR 2 = 3.
 * 3) 5 XOR 2 = 7.
 *
 * Example 2:
 *
 * Input: nums = [5,2,4,6,6,3], queries = [[12,4],[8,1],[6,3]]
 * Output: [15,-1,5]
 *
 *  
 * Constraints:
 *
 * 	1 <= nums.length, queries.length <= 10^5
 * 	queries[i].length == 2
 * 	0 <= nums[j], xi, mi <= 10^9
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/maximum-xor-with-an-element-from-array/
// discuss: https://leetcode.com/problems/maximum-xor-with-an-element-from-array/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

#[derive(Debug, Clone, Default)]
struct Node {
    children: [Option<Box<Node>>; 2],
}

impl Node {
    pub fn insert(&mut self, n: i32) {
        do_insert(self, n, 31);
    }
}

fn do_insert(node: &mut Node, num: i32, bit: u32) {
    let b = ((num & (1 << bit)) != 0) as usize;
    if node.children[b].is_none() {
        node.children[b] = Some(Box::new(Node::default()));
    }

    if bit > 0 {
        do_insert(node.children[b].as_mut().unwrap(), num, bit - 1);
    }
}

impl Solution {
    // Credit: https://leetcode.com/problems/maximum-xor-with-an-element-from-array/solutions/1762758/rust-trie-solution/
    pub fn maximize_xor(nums: Vec<i32>, queries: Vec<Vec<i32>>) -> Vec<i32> {
        let mut result_indexes = std::collections::HashMap::new();
        for (idx, query) in queries.iter().enumerate() {
            result_indexes
                .entry((query[0], query[1]))
                .or_insert(vec![])
                .push(idx);
        }

        let mut nums = nums.clone();
        let mut queries = queries.clone();

        nums.sort_unstable();
        queries.sort_unstable_by(|a, b| a[1].cmp(&b[1]));

        let mut result = vec![0; queries.len()];
        let mut trie = Node::default();

        let mut next_idx = 0;

        let mut last_query = None;

        for query in queries.iter() {
            if Some(query) == last_query {
                continue;
            }
            last_query = Some(query);

            let val = query[0];
            let lim = query[1];

            for idx in next_idx..nums.len() {
                let n = nums[idx];
                if n > lim {
                    break;
                }

                trie.insert(n);
                next_idx = idx + 1;
            }

            let mut xored = -1;
            if next_idx != 0 {
                let mut node = &trie;
                xored = 0;

                for bit in (0..32).rev() {
                    let is_set = (val & (1 << bit)) != 0;
                    match &node.children[(is_set ^ true) as usize] {
                        None => {
                            node = node.children[is_set as usize].as_ref().unwrap();
                        }

                        Some(child) => {
                            xored |= 1 << bit;
                            node = child;
                        }
                    }
                }
            }

            for &idx in result_indexes.get(&(val, lim)).unwrap() {
                result[idx] = xored;
            }
        }

        result
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_1707_example_1() {
        let nums = vec![0, 1, 2, 3, 4];
        let queries = vec![vec![3, 1], vec![1, 3], vec![5, 6]];

        let result = vec![3, 3, 7];

        assert_eq!(Solution::maximize_xor(nums, queries), result);
    }

    #[test]
    fn test_1707_example_2() {
        let nums = vec![5, 2, 4, 6, 6, 3];
        let queries = vec![vec![12, 4], vec![8, 1], vec![6, 3]];

        let result = vec![15, -1, 5];

        assert_eq!(Solution::maximize_xor(nums, queries), result);
    }
}

// Accepted solution for LeetCode #1707: Maximum XOR With an Element From Array
class Trie {
    children: (Trie | null)[];

    constructor() {
        this.children = [null, null];
    }

    insert(x: number): void {
        let node: Trie | null = this;
        for (let i = 30; ~i; i--) {
            const v = (x >> i) & 1;
            if (node.children[v] === null) {
                node.children[v] = new Trie();
            }
            node = node.children[v] as Trie;
        }
    }

    search(x: number): number {
        let node: Trie | null = this;
        let ans = 0;
        for (let i = 30; ~i; i--) {
            const v = (x >> i) & 1;
            if (node.children[v ^ 1] !== null) {
                ans |= 1 << i;
                node = node.children[v ^ 1] as Trie;
            } else if (node.children[v] !== null) {
                node = node.children[v] as Trie;
            } else {
                return -1;
            }
        }
        return ans;
    }
}

function maximizeXor(nums: number[], queries: number[][]): number[] {
    nums.sort((a, b) => a - b);
    const n = queries.length;
    const idx = Array.from({ length: n }, (_, i) => i);
    idx.sort((i, j) => queries[i][1] - queries[j][1]);
    const ans: number[] = [];
    const trie = new Trie();
    let j = 0;
    for (const i of idx) {
        const x = queries[i][0];
        const m = queries[i][1];
        while (j < nums.length && nums[j] <= m) {
            trie.insert(nums[j++]);
        }
        ans[i] = trie.search(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(m × log m + n × (log n + log M)

Space

O(n × log M)

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.