Maximum XOR of Two Numbers in an Array - Practice Coding Problems

Maximum XOR of Two Numbers in an Array - Problem

Given an integer array nums, your task is to find the maximum XOR (exclusive OR) value that can be obtained by combining any two numbers from the array.

The XOR operation compares bits of two numbers and returns 1 when bits are different, 0 when they're the same. For example: 3 XOR 5 = 011 XOR 101 = 110 = 6.

Goal: Return the maximum result of nums[i] XOR nums[j] where 0 ≤ i ≤ j < n.

Note: You can use the same element twice (when i = j) or different elements.

Input & Output

example_1.py — Basic case

$ Input: nums = [3, 10, 5, 25, 2, 8]

› Output: 28

💡 Note: The maximum result is 5 XOR 25 = 28. In binary: 101 XOR 11001 = 11100 = 28

example_2.py — Small array

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

› Output: 127

💡 Note: The maximum XOR is achieved by combining 36 (100100) and 91 (1011011) = 127

example_3.py — Edge case

$ Input: nums = [8, 10, 2]

› Output: 10

💡 Note: Possible XORs: 8^10=2, 8^2=10, 10^2=8. Maximum is 10

Visualization

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Understanding the Visualization

Build Binary Tree

Create a Trie storing all numbers as binary paths from most significant bit

Seek Opposites

For each number, traverse the tree always trying to go the opposite direction

Maximize Each Bit

When possible, choose the path that makes the current bit position 1 in the XOR result

Track Maximum

Keep track of the highest XOR value found across all number combinations

Key Takeaway

🎯 Key Insight: XOR is maximized when bits differ in the most significant positions. The Trie allows us to greedily choose the best partner for each number by always seeking opposite bits.

Time & Space Complexity

Time Complexity

⏱️

O(n²)

We check all possible pairs of numbers, which gives us n×(n-1)/2 ≈ n² comparisons

⚠ Quadratic Growth

Space Complexity

O(1)

Only using a few variables to track the maximum XOR value

✓ Linear Space

Constraints

1 ≤ nums.length ≤ 2 × 10⁴
0 ≤ nums[i] ≤ 2³¹ - 1
Follow up: Can you solve it in O(n log(max_value)) time?

Asked in

G Google 45 a Amazon 38 f Meta 32 ⊞ Microsoft 28 Apple 22

The optimal solution uses a Trie (prefix tree) to store binary representations of all numbers. For each number, we traverse the Trie trying to take the opposite bit path at each level to maximize XOR. This achieves O(n × log(max_val)) time complexity, much better than the O(n²) brute force approach.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Nested Loops)	O(n²)	O(1)	Check all possible pairs of numbers to find maximum XOR
Bit-by-Bit Construction (Optimal)	O(n × log(max_val))	O(n × log(max_val))	Build the maximum XOR bit by bit using a Trie data structure

Brute Force (Nested Loops) — Algorithm Steps

Initialize maxXor to 0
Use nested loops to check all pairs (i,j) where i ≤ j
For each pair, compute nums[i] XOR nums[j]
Update maxXor if current XOR is larger
Return the maximum XOR found

Visualization

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Step-by-Step Walkthrough

Initialize

Start with maxXor = 0

Compare Pairs

Check each pair (i,j) and compute XOR

Update Maximum

Keep track of the largest XOR value found

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>

int findMaximumXOR(int* nums, int numsSize) {
    int maxXor = 0;
    
    for (int i = 0; i < numsSize; i++) {
        for (int j = i; j < numsSize; j++) {
            int currentXor = nums[i] ^ nums[j];
            if (currentXor > maxXor) {
                maxXor = currentXor;
            }
        }
    }
    
    return maxXor;
}

int main() {
    int nums[] = {3, 10, 5, 25, 2, 8};
    int size = sizeof(nums) / sizeof(nums[0]);
    printf("%d\n", findMaximumXOR(nums, size));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

We check all possible pairs of numbers, which gives us n×(n-1)/2 ≈ n² comparisons

⚠ Quadratic Growth

Space Complexity

O(1)

Only using a few variables to track the maximum XOR value

✓ Linear Space

Constraints

1 ≤ nums.length ≤ 2 × 10⁴
0 ≤ nums[i] ≤ 2³¹ - 1
Follow up: Can you solve it in O(n log(max_value)) time?

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Input & Output

Visualization

Time & Space Complexity

Related Problems

Constraints

Common Approaches

Brute Force (Nested Loops) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Constraints

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