What is the time complexity of Find the Difference of Two Arrays?

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

What is the space complexity of Find the Difference of Two Arrays?

The space complexity is O(n + m).

What data structures are used to solve Find the Difference of Two Arrays?

This problem uses Array, Hash Table. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Find the Difference of Two Arrays?

Using sets to efficiently identify unique elements and check for membership: Sets provide constant-time average complexity for `add`, `contains`, and `remove` operations, making the overall process significantly faster than nested loops. Leveraging set difference operations: Set difference (`-` operator in Python, `removeAll` in Java, etc.) directly computes the elements present in one set but not the other, simplifying the logic.

Which companies ask LeetCode 2215. Find the Difference of Two Arrays in interviews?

This problem is frequently asked at Yandex.

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Find the Difference of Two Arrays - LeetCode 2215 Solution

Find the Difference of Two Arrays - Complete Solution Guide

Find the Difference of Two Arrays is LeetCode problem 2215, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given two 0-indexed integer arrays nums1 and nums2 , return a list answer of size 2 where: answer[0] is a list of all distinct integers in nums1 which are not present in nums2 . answer[1] is a list of all distinct integers in nums2 which are not present in nums1 . Note that the integers in the lists may be returned in any order. Example 1: Input: nums1 = [1,2,3], nums2 = [2,4,6] Output: [[1,3],[4,6]] Explanation: For nums1, nums1[1] = 2 is present at index 0 of nums2, whereas nums1[0] = 1 and nu

Detailed Explanation

The problem asks to find the difference between two arrays, `nums1` and `nums2`. The output should be a list containing two sub-lists. The first sub-list contains unique elements present in `nums1` but not in `nums2`. The second sub-list contains unique elements present in `nums2` but not in `nums1`. The order of elements within each sub-list doesn't matter. The input arrays can contain duplicate elements, but the output arrays must only contain unique elements. Both input arrays contain integers within a specified range.

Solution Approach

The provided solutions utilize sets (or hash sets) to efficiently solve this problem. First, the input arrays are converted into sets to eliminate duplicate elements. Then, set difference operations are used to find the unique elements present in one array but not the other. Finally, these unique elements are converted back into lists and returned as a list of lists.

Step-by-Step Algorithm

Convert `nums1` and `nums2` into sets (`set1` and `set2`). This automatically removes duplicates.
Compute the set difference `set1 - set2` to find elements in `nums1` but not in `nums2`. Convert this result into a list.
Compute the set difference `set2 - set1` to find elements in `nums2` but not in `nums1`. Convert this result into a list.
Return a list containing the two lists obtained in steps 2 and 3.

Key Insights

Using sets to efficiently identify unique elements and check for membership: Sets provide constant-time average complexity for `add`, `contains`, and `remove` operations, making the overall process significantly faster than nested loops.
Leveraging set difference operations: Set difference (`-` operator in Python, `removeAll` in Java, etc.) directly computes the elements present in one set but not the other, simplifying the logic.
Handling duplicates: The problem explicitly requires handling duplicate numbers in input arrays. Sets inherently address this by storing only unique elements.

Complexity Analysis

Time Complexity: O(n + m)

Space Complexity: O(n + m)

Topics

This problem involves: Array, Hash Table.

Companies

Asked at: Yandex.

Find the Difference of Two Arrays | LeetCode 2215

Easy

Array Hash Table

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

Solution Code

Available in:

import java.util.List;
import java.util.ArrayList;
import java.util.Set;
import java.util.HashSet;

class Solution {
    public List<List<Integer>> findDifference(int[] nums1, int[] nums2) {
        Set<Integer> set1 = new HashSet<>();
        Set<Integer> set2 = new HashSet<>();

        for (int num : nums1) {
            set1.add(num);
        }
        for (int num : nums2) {
            set2.add(num);
        }

        List<Integer> list1 = new ArrayList<>();
        for (int num : nums1) {
            if (!set2.contains(num)) {
                if (!list1.contains(num)) {
                    list1.add(num);
                }
            }
        }

        List<Integer> list2 = new ArrayList<>();
        for (int num : nums2) {
            if (!set1.contains(num)) {
                if (!list2.contains(num)) {
                    list2.add(num);
                }
            }
        }

        List<List<Integer>> result = new ArrayList<>();
        result.add(list1);
        result.add(list2);

        return result;
    }
}

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Problem Info

Description

Problem 2215: Find the Difference of Two Arrays - Detailed explanation and solution approaches.

Topics

Array Hash Table

Companies

Yandex

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:2215

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
Convert `nums1` and `nums2` into sets (`set1` and `set2`). This automatically removes duplicates.
2
Compute the set difference `set1 - set2` to find elements in `nums1` but not in `nums2`. Convert this result into a list.
3
Compute the set difference `set2 - set1` to find elements in `nums2` but not in `nums1`. Convert this result into a list.
4
Return a list containing the two lists obtained in steps 2 and 3.

💡

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(n + m)

📝

Why?

Creating sets takes O(m) and O(n) time respectively, where 'm' and 'n' are the lengths of `nums1` and `nums2`. Set difference operations also take O(m) and O(n) on average. Converting sets back to lists takes linear time. Therefore, the overall time complexity is linear with respect to the input size. The space complexity is also linear because the size of the sets is bounded by 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

Nested Loops Approach

This approach uses nested loops to iterate through each array and check for the presence of each element in the other array. It's less efficient.

Time: O(m*n)Space: O(1) or O(m+n) depending on implementation (if temporary lists are used)

Trade-offs: Simpler to understand initially but far less efficient than the set-based approach for larger inputs. The set-based approach is significantly better for larger input sizes, while this approach might be acceptable for very small arrays where readability is prioritized over performance.

✨

Key Insights

1Using sets to efficiently identify unique elements and check for membership: Sets provide constant-time average complexity for `add`, `contains`, and `remove` operations, making the overall process significantly faster than nested loops.
2Leveraging set difference operations: Set difference (`-` operator in Python, `removeAll` in Java, etc.) directly computes the elements present in one set but not the other, simplifying the logic.
3Handling duplicates: The problem explicitly requires handling duplicate numbers in input arrays. Sets inherently address this by storing only unique elements.

📚

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

Empty input arrays: The code handles empty input arrays gracefully; the set difference will result in empty sets, which are correctly converted into empty lists.
Arrays with all duplicate elements: The use of sets automatically handles this case. If both arrays contain only duplicates, the result would be two empty arrays.

⚠️

Common Mistakes to Avoid

✗Forgetting to handle duplicates: A common mistake is to not eliminate duplicates before comparing elements. This can lead to incorrect results.

✗Inefficiently checking for element presence: Using nested loops to check for element presence has O(m*n) time complexity instead of O(n+m) using sets.

✅

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?

💼Data comparison and cleaning: Identifying differences in two datasets (e.g., comparing customer databases to identify new or deleted accounts).

🏦Version control systems: Tracking changes between different versions of code or data by finding the unique elements added or removed.

💡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

Use this problem to demonstrate clean code practices. Variable naming and code organization matter.

⚠️ Watch Out

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

👨‍💻 From Experience

This is one of the first problems I solved when learning Array. It's a great foundation for more complex problems.

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: July 26, 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 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.

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

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 "Find the Difference of Two Arrays"?

For this easy-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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