LeetCode #421 — MEDIUM

Maximum XOR of Two Numbers in an Array

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

The Problem

Problem Statement

Given an integer array nums, return the maximum result of nums[i] XOR nums[j], where 0 <= i <= j < n.

Example 1:

Input: nums = [3,10,5,25,2,8]
Output: 28
Explanation: The maximum result is 5 XOR 25 = 28.

Example 2:

Input: nums = [14,70,53,83,49,91,36,80,92,51,66,70]
Output: 127

Constraints:

1 <= nums.length <= 2 * 10⁵
0 <= nums[i] <= 2³¹ - 1

Step 01

Brute Force Baseline

Problem summary: Given an integer array nums, return the maximum result of nums[i] XOR nums[j], where 0 <= i <= j < n.

Baseline thinking

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

Pattern signal: Array · Hash Map · Bit Manipulation · Trie

Example 1

[3,10,5,25,2,8]

Example 2

[14,70,53,83,49,91,36,80,92,51,66,70]

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

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 #421: Maximum XOR of Two Numbers in an Array
class Trie {
    private Trie[] children = new Trie[2];

    public Trie() {
    }

    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 {
                node = node.children[v];
            }
        }
        return ans;
    }
}

class Solution {
    public int findMaximumXOR(int[] nums) {
        Trie trie = new Trie();
        int ans = 0;
        for (int x : nums) {
            trie.insert(x);
            ans = Math.max(ans, trie.search(x));
        }
        return ans;
    }
}

// Accepted solution for LeetCode #421: Maximum XOR of Two Numbers in an 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 {
			node = node.children[v]
		}
	}
	return ans
}

func findMaximumXOR(nums []int) (ans int) {
	trie := newTrie()
	for _, x := range nums {
		trie.insert(x)
		ans = max(ans, trie.search(x))
	}
	return ans
}

# Accepted solution for LeetCode #421: Maximum XOR of Two Numbers in an Array
class Trie:
    __slots__ = ("children",)

    def __init__(self):
        self.children: List[Trie | None] = [None, None]

    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]
            else:
                node = node.children[v]
        return ans


class Solution:
    def findMaximumXOR(self, nums: List[int]) -> int:
        trie = Trie()
        for x in nums:
            trie.insert(x)
        return max(trie.search(x) for x in nums)

// Accepted solution for LeetCode #421: Maximum XOR of Two Numbers in an Array
struct Trie {
    children: [Option<Box<Trie>>; 2],
}

impl Trie {
    fn new() -> Trie {
        Trie {
            children: [None, None],
        }
    }

    fn insert(&mut self, x: i32) {
        let mut node = self;
        for i in (0..=30).rev() {
            let v = ((x >> i) & 1) as usize;
            if node.children[v].is_none() {
                node.children[v] = Some(Box::new(Trie::new()));
            }
            node = node.children[v].as_mut().unwrap();
        }
    }

    fn search(&self, x: i32) -> i32 {
        let mut node = self;
        let mut ans = 0;
        for i in (0..=30).rev() {
            let v = ((x >> i) & 1) as usize;
            if let Some(child) = &node.children[v ^ 1] {
                ans |= 1 << i;
                node = child.as_ref();
            } else {
                node = node.children[v].as_ref().unwrap();
            }
        }
        ans
    }
}

impl Solution {
    pub fn find_maximum_xor(nums: Vec<i32>) -> i32 {
        let mut trie = Trie::new();
        let mut ans = 0;
        for &x in nums.iter() {
            trie.insert(x);
            ans = ans.max(trie.search(x));
        }
        ans
    }
}

// Accepted solution for LeetCode #421: Maximum XOR of Two Numbers in an Array
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #421: Maximum XOR of Two Numbers in an Array
// class Trie {
//     private Trie[] children = new Trie[2];
// 
//     public Trie() {
//     }
// 
//     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 {
//                 node = node.children[v];
//             }
//         }
//         return ans;
//     }
// }
// 
// class Solution {
//     public int findMaximumXOR(int[] nums) {
//         Trie trie = new Trie();
//         int ans = 0;
//         for (int x : nums) {
//             trie.insert(x);
//             ans = Math.max(ans, 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(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.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.