What is the time complexity of Maximum XOR of Two Numbers in an Array?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Maximum XOR of Two Numbers in an Array?

This problem uses Array, Hash Table, Bit Manipulation, Trie. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Maximum XOR of Two Numbers in an Array?

Insight 1: Building the result bit by bit from the most significant bit (MSB) downwards allows us to greedily maximize the XOR sum. Insight 2: Using a set (or hash table) to store prefixes of numbers enables efficient checking if a complementary prefix exists that would increase the XOR sum.

Which companies ask LeetCode 421. Maximum XOR of Two Numbers in an Array in interviews?

This problem is frequently asked at Goldman Sachs.

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Maximum XOR of Two Numbers in an Array - LeetCode 421 Solution

Q: How to solve Maximum XOR of Two Numbers in an Array?

The solution utilizes a greedy approach combined with bit manipulation and a hash set. It starts from the most significant bit (MSB) and iterates down to the least significant bit. In each iteration, it tries to maximize the XOR sum by setting the current bit to 1. It checks if there exist two numbers in the array whose prefixes (up to the current bit) XOR to the desired target, which is `max_xor | (1 << i)`. If such a pair exists, it means setting the current bit to 1 is possible, and the `max_xor` is updated accordingly.

Maximum XOR of Two Numbers in an Array - Complete Solution Guide

Maximum XOR of Two Numbers in an Array is LeetCode problem 421, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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 * 10 5 0 <= nums[i] <= 2 31 - 1

Detailed Explanation

The problem asks us to find the maximum XOR value that can be obtained by XORing any two numbers in a given array of integers. The XOR operation compares the bits of two numbers. If the bits are different, the result is 1; otherwise, the result is 0. The goal is to iterate through all possible pairs of numbers in the array and find the pair that yields the largest XOR value.

Solution Approach

The solution utilizes a greedy approach combined with bit manipulation and a hash set. It starts from the most significant bit (MSB) and iterates down to the least significant bit. In each iteration, it tries to maximize the XOR sum by setting the current bit to 1. It checks if there exist two numbers in the array whose prefixes (up to the current bit) XOR to the desired target, which is `max_xor | (1 << i)`. If such a pair exists, it means setting the current bit to 1 is possible, and the `max_xor` is updated accordingly.

Step-by-Step Algorithm

Step 1: Initialize `max_xor` to 0 and `mask` to 0. `max_xor` will store the maximum XOR sum found so far, and `mask` will be used to extract the prefixes of the numbers.
Step 2: Iterate from the 30th bit (MSB) down to the 0th bit. This loop considers each bit position from the most significant to the least significant.
Step 3: Update the `mask` to include the current bit. This is done by performing a bitwise OR operation: `mask |= (1 << i)`. The mask will isolate the bits from the MSB down to the current bit `i`.
Step 4: Create a set called `prefixes` and store the prefixes of all numbers in the array. The prefix of a number is obtained by performing a bitwise AND operation with the `mask`: `num & mask`.
Step 5: Create a temporary variable `temp` and set the current bit of `max_xor` to 1: `temp = max_xor | (1 << i)`. This represents the potential maximum XOR sum if the current bit is set to 1.
Step 6: Iterate through the `prefixes` set. For each prefix `p`, check if `(temp ^ p)` also exists in the `prefixes` set. If it does, it means there exists another number in the array whose prefix XORed with `p` equals `temp`. This indicates that setting the current bit to 1 is possible, and `max_xor` is updated to `temp`.
Step 7: If a valid XOR is found (step 6), break the inner loop. Otherwise, continue iterating through the `prefixes` set to find a suitable pair.
Step 8: After iterating through all bits, return the `max_xor` value.

Key Insights

Insight 1: Building the result bit by bit from the most significant bit (MSB) downwards allows us to greedily maximize the XOR sum.
Insight 2: Using a set (or hash table) to store prefixes of numbers enables efficient checking if a complementary prefix exists that would increase the XOR sum.
Insight 3: The constraint that each number is at most 2<sup>31</sup> - 1 implies that we need to consider 31 bits (bits 0 to 30) to represent the maximum possible XOR value.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, Hash Table, Bit Manipulation, Trie.

Companies

Asked at: Goldman Sachs.

Maximum XOR of Two Numbers in an Array | LeetCode 421

Medium

Array Hash Table Bit Manipulation Trie

Time: O(n)Space: O(n)

Solution Code

Available in:

import java.util.HashSet;
import java.util.Set;

class Solution {
    public int findMaximumXOR(int[] nums) {
        int max_xor = 0;
        int mask = 0;
        for (int i = 30; i >= 0; i--) {
            mask |= (1 << i);

            Set<Integer> prefixes = new HashSet<>();
            for (int num : nums) {
                prefixes.add(num & mask);
            }

            int temp = max_xor | (1 << i);

            for (int p : prefixes) {
                if (prefixes.contains(temp ^ p)) {
                    max_xor = temp;
                    break;
                }
            }

        }
        return max_xor;
    }
}

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Problem Info

Description

Problem 421: Maximum XOR of Two Numbers in an Array - Detailed explanation and solution approaches.

Topics

Array Hash Table Bit Manipulation Trie

Companies

Goldman Sachs

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:421

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

8 steps

1
Step 1: Initialize `max_xor` to 0 and `mask` to 0. `max_xor` will store the maximum XOR sum found so far, and `mask` will be used to extract the prefixes of the numbers.
2
Step 2: Iterate from the 30th bit (MSB) down to the 0th bit. This loop considers each bit position from the most significant to the least significant.
3
Step 3: Update the `mask` to include the current bit. This is done by performing a bitwise OR operation: `mask |= (1 << i)`. The mask will isolate the bits from the MSB down to the current bit `i`.
4
Step 4: Create a set called `prefixes` and store the prefixes of all numbers in the array. The prefix of a number is obtained by performing a bitwise AND operation with the `mask`: `num & mask`.
5
Step 5: Create a temporary variable `temp` and set the current bit of `max_xor` to 1: `temp = max_xor | (1 << i)`. This represents the potential maximum XOR sum if the current bit is set to 1.
6
Step 6: Iterate through the `prefixes` set. For each prefix `p`, check if `(temp ^ p)` also exists in the `prefixes` set. If it does, it means there exists another number in the array whose prefix XORed with `p` equals `temp`. This indicates that setting the current bit to 1 is possible, and `max_xor` is updated to `temp`.
7
Step 7: If a valid XOR is found (step 6), break the inner loop. Otherwise, continue iterating through the `prefixes` set to find a suitable pair.
8
Step 8: After iterating through all bits, return the `max_xor` value.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is dominated by the loop that iterates over the bits and the creation and iteration over prefixes, which are both proportional to n. The space complexity is due to the prefix set, which can grow to the size of the input array in the worst case.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Trie-based solution

A Trie (prefix tree) can be used to store the binary representation of each number in the array. Then, for each number, the Trie can be traversed to find the number that maximizes the XOR value.

Time: O(n), where n is the number of elements in the input array `nums`. Each insertion and search in the Trie takes O(31) time, as we have to traverse up to 31 levels of the tree.Space: O(n), where n is the number of elements in the input array `nums`. In the worst case, the Trie will store all the numbers in the array, each with 31 bits.

Trade-offs: The Trie-based solution offers a slightly different perspective but has the same asymptotic time and space complexity. It might be preferred in situations where frequent XOR maximization operations are required, as the Trie can be precomputed and reused.

✨

Key Insights

1Insight 1: Building the result bit by bit from the most significant bit (MSB) downwards allows us to greedily maximize the XOR sum.
2Insight 2: Using a set (or hash table) to store prefixes of numbers enables efficient checking if a complementary prefix exists that would increase the XOR sum.
3Insight 3: The constraint that each number is at most 2<sup>31</sup> - 1 implies that we need to consider 31 bits (bits 0 to 30) to represent the maximum possible XOR value.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Edge Cases

Edge case 1: Empty array. While the problem constraints say the array length is at least 1, good practice to consider this. The code will still technically function correctly because the loops won't execute and it will return 0.
Edge case 2: Array with a single element. Similar to above, the loops won't execute and the function returns 0, which is technically correct as the XOR of a number with itself is 0.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not considering all 31 bits. It's important to iterate from 30 down to 0 to ensure that all possible XOR values are considered.

✗Common mistake 2: Incorrectly creating the mask. Ensure the mask isolates the correct bits by using the bitwise OR and left shift operations appropriately.

✗Common mistake 3: Forgetting to handle the case where no suitable XOR pair is found for a given bit. The code should proceed to the next bit without prematurely updating `max_xor`.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Network routing protocols. XOR operations are used in routing tables to efficiently determine the next hop for a packet.

🏦Real-world scenario 2: Cryptography. XOR is a fundamental operation used in many encryption algorithms due to its reversibility. Finding maximum XOR values might be used in specific cryptographic analyses or optimizations.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: August 1, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 13

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 6

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Maximum XOR of Two Numbers in an Array"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether hash table techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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