What is the time complexity of Bitwise XOR of All Pairings?

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

What data structures are used to solve Bitwise XOR of All Pairings?

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

What are the key insights for solving Bitwise XOR of All Pairings?

Insight 1: The XOR operation is associative and commutative, meaning the order in which we perform XOR operations doesn't matter, and we can group them as needed. Insight 2: The XOR of a number with itself is 0. If a number appears an even number of times in a series of XOR operations, it effectively cancels itself out.

Which companies ask LeetCode 2425. Bitwise XOR of All Pairings in interviews?

This problem is frequently asked at Trilogy.

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Bitwise XOR of All Pairings - LeetCode 2425 Solution

Bitwise XOR of All Pairings - Complete Solution Guide

Bitwise XOR of All Pairings is LeetCode problem 2425, 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

You are given two 0-indexed arrays, nums1 and nums2 , consisting of non-negative integers. Let there be another array, nums3 , which contains the bitwise XOR of all pairings of integers between nums1 and nums2 (every integer in nums1 is paired with every integer in nums2 exactly once ). Return the bitwise XOR of all integers in nums3 . Example 1: Input: nums1 = [2,1,3], nums2 = [10,2,5,0] Output: 13 Explanation: A possible nums3 array is [8,0,7,2,11,3,4,1,9,1,6,3]. The bitwise XOR of all these n

Detailed Explanation

The problem asks us to calculate the bitwise XOR of all possible pairings of elements from two input arrays, `nums1` and `nums2`. Specifically, for each element in `nums1`, we XOR it with every element in `nums2`, creating a new array `nums3` containing all these XOR results. The goal is to find the bitwise XOR of all the numbers in `nums3`. The input arrays `nums1` and `nums2` contain non-negative integers, and their lengths are up to 10^5.

Solution Approach

The solution avoids constructing `nums3` entirely. Instead, it leverages the properties of the XOR operation. If the length of `nums2` is odd, each element in `nums1` will be XORed an odd number of times (specifically, `nums2.length` times). Therefore, we need to XOR all the elements of `nums1`. If the length of `nums2` is even, each element in `nums1` is XORed an even number of times and will therefore cancel out. A similar logic applies to `nums2` based on the length of `nums1`.

Step-by-Step Algorithm

Step 1: Initialize a variable `res` to 0. This will store the final XOR result.
Step 2: Check if the length of `nums2` (`m`) is odd. If it is, iterate through `nums1` and XOR each element with `res`.
Step 3: Check if the length of `nums1` (`n`) is odd. If it is, iterate through `nums2` and XOR each element with `res`.
Step 4: Return the final value of `res`.

Key Insights

Insight 1: The XOR operation is associative and commutative, meaning the order in which we perform XOR operations doesn't matter, and we can group them as needed.
Insight 2: The XOR of a number with itself is 0. If a number appears an even number of times in a series of XOR operations, it effectively cancels itself out.
Insight 3: We don't need to explicitly construct `nums3`. We can determine the XOR result directly based on the lengths of `nums1` and `nums2` and the XOR of their elements.

Complexity Analysis

Time Complexity: O(n + m)

Space Complexity: O(1)

Topics

This problem involves: Array, Bit Manipulation, Brainteaser.

Companies

Asked at: Trilogy.

Bitwise XOR of All Pairings | LeetCode 2425

Medium

Array Bit Manipulation Brainteaser

Time: O(n + m)Space: O(1)

Solution Code

Available in:

class Solution {
    public int xorAllNums(int[] nums1, int[] nums2) {
        int n = nums1.length;
        int m = nums2.length;
        
        int res = 0;
        
        if (m % 2 != 0) {
            for (int num : nums1) {
                res ^= num;
            }
        }
        
        if (n % 2 != 0) {
            for (int num : nums2) {
                res ^= num;
            }
        }
                
        return res;
    }
}

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Problem Info

Description

Problem 2425: Bitwise XOR of All Pairings - Detailed explanation and solution approaches.

Topics

Array Bit Manipulation Brainteaser

Companies

Trilogy

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2425

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

4 steps

1
Step 1: Initialize a variable `res` to 0. This will store the final XOR result.
2
Step 2: Check if the length of `nums2` (`m`) is odd. If it is, iterate through `nums1` and XOR each element with `res`.
3
Step 3: Check if the length of `nums1` (`n`) is odd. If it is, iterate through `nums2` and XOR each element with `res`.
4
Step 4: Return the final value of `res`.

💡

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 + m)

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is determined by the two loops, each iterating through one of the input arrays. The space complexity is constant since we only use a few integer variables regardless of the input array sizes.

📖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

Brute Force

Construct `nums3` explicitly by XORing every element of `nums1` with every element of `nums2`, then XOR all the elements of `nums3` together.

Time: O(n * m), where n is the length of `nums1` and m is the length of `nums2`. This is because constructing `nums3` requires n * m XOR operations, and XORing all its elements takes another n * m - 1 operations which still equates to O(n * m).Space: O(n * m), to store `nums3`.

Trade-offs: The brute force approach is simpler to understand initially but significantly less efficient than the optimized solution, especially for large arrays. It uses more memory and has a higher time complexity, making it impractical for the problem's constraints.

✨

Key Insights

1Insight 1: The XOR operation is associative and commutative, meaning the order in which we perform XOR operations doesn't matter, and we can group them as needed.
2Insight 2: The XOR of a number with itself is 0. If a number appears an even number of times in a series of XOR operations, it effectively cancels itself out.
3Insight 3: We don't need to explicitly construct `nums3`. We can determine the XOR result directly based on the lengths of `nums1` and `nums2` and the XOR of their elements.

📚

Key Terms Explained

For Students

Array

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

Edge Cases

Edge case 1: One or both of the input arrays are empty. The algorithm correctly handles this case because if an array is empty, its corresponding loop will not execute, and `res` will either remain 0 or reflect the XOR of the other array's elements as required.
Edge case 2: Arrays contain very large numbers. The solution utilizes bitwise XOR, which is suitable for large numbers as long as they fit within the integer limits of the chosen programming language. No special handling is needed.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Attempting to create the `nums3` array explicitly, leading to excessive memory usage and a slower solution.

✗Common mistake 2: Overlooking the properties of the XOR operation and trying to solve the problem with more complex and unnecessary logic.

✅

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: Data integrity checks. XOR is used to create checksums for data transmission or storage. By XORing all the data bits, a checksum is generated. If any bit is flipped during transmission or storage, the checksum will change, indicating data corruption.

🏦Real-world scenario 2: Cryptography. XOR is used in simple encryption algorithms. Combining a message with a key using XOR creates an encrypted message. XORing the encrypted message with the same key decrypts it back to the original message.

💡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.

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

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 25, 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

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 13

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

@AlgoMasterJul 6

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

@InterviewProJun 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 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 "Bitwise XOR of All Pairings"?

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 bit manipulation techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?