What is the time complexity of Concatenation of Array?

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

What data structures are used to solve Concatenation of Array?

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

What are the key insights for solving Concatenation of Array?

Insight 1: The problem's core is simple array manipulation. No complex data structures or algorithms are needed. Insight 2: Directly creating a new array of double the size and copying elements is a straightforward and efficient approach.

Which companies ask LeetCode 1929. Concatenation of Array in interviews?

This problem is frequently asked at GE Healthcare.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Concatenation of Array - LeetCode 1929 Solution

Q: How to solve Concatenation of Array?

The provided solutions all use the same approach: They create a new array `ans` twice the size of the input array `nums`. Then, they iterate through `nums`, copying each element into `ans` twice – once in the first half and once in the second half of `ans`. This direct approach is very intuitive and easy to understand.

Concatenation of Array - Complete Solution Guide

Concatenation of Array is LeetCode problem 1929, 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 an integer array nums of length n , you want to create an array ans of length 2n where ans[i] == nums[i] and ans[i + n] == nums[i] for 0 <= i < n ( 0-indexed ). Specifically, ans is the concatenation of two nums arrays. Return the array ans . Example 1: Input: nums = [1,2,1] Output: [1,2,1,1,2,1] Explanation: The array ans is formed as follows: - ans = [nums[0],nums[1],nums[2],nums[0],nums[1],nums[2]] - ans = [1,2,1,1,2,1] Example 2: Input: nums = [1,3,2,1] Output: [1,3,2,1,1,3,2,1] Explan

Detailed Explanation

The problem asks you to take an integer array `nums` and create a new array `ans` that is twice the length of `nums`. The first half of `ans` should be identical to `nums`, and the second half should also be identical to `nums`. In essence, you're concatenating `nums` with itself to create `ans`. The input is an integer array, and the output is a new integer array that's the concatenation of the input array with itself. Constraints limit the input array's length and the values within the array.

Solution Approach

The provided solutions all use the same approach: They create a new array `ans` twice the size of the input array `nums`. Then, they iterate through `nums`, copying each element into `ans` twice – once in the first half and once in the second half of `ans`. This direct approach is very intuitive and easy to understand.

Step-by-Step Algorithm

Step 1: Determine the length of the input array `nums` and store it in a variable `n`.
Step 2: Create a new array `ans` with a size of `2 * n`.
Step 3: Iterate from `i = 0` to `n - 1`. In each iteration:
Step 4: Assign `nums[i]` to `ans[i]` (copying to the first half).
Step 5: Assign `nums[i]` to `ans[i + n]` (copying to the second half).
Step 6: Return the `ans` array.

Key Insights

Insight 1: The problem's core is simple array manipulation. No complex data structures or algorithms are needed.
Insight 2: Directly creating a new array of double the size and copying elements is a straightforward and efficient approach.
Insight 3: The problem highlights the importance of understanding array indexing and iterating through arrays.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, Simulation.

Companies

Asked at: GE Healthcare.

Concatenation of Array | LeetCode 1929

Easy

Array Simulation

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

Solution Code

Available in:

class Solution {
    public int[] getConcatenation(int[] nums) {
        int n = nums.length;
        int[] ans = new int[2 * n];
        for (int i = 0; i < n; i++) {
            ans[i] = nums[i];
            ans[i + n] = nums[i];
        }
        return ans;
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 1929: Concatenation of Array - Detailed explanation and solution approaches.

Topics

Array Simulation

Companies

GE Healthcare

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:1929

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

6 steps

1
Step 1: Determine the length of the input array `nums` and store it in a variable `n`.
2
Step 2: Create a new array `ans` with a size of `2 * n`.
3
Step 3: Iterate from `i = 0` to `n - 1`. In each iteration:
4
Step 4: Assign `nums[i]` to `ans[i]` (copying to the first half).
5
Step 5: Assign `nums[i]` to `ans[i + n]` (copying to the second half).
6
Step 6: Return the `ans` array.

💡

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 O(n) because the code iterates through the input array `nums` once. The space complexity is O(n) because a new array `ans` is created with a size proportional to the size of `nums`.

📖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

Using `extend` (Python)

In Python, you could use the `extend` method to concatenate the list with itself directly: `ans = nums.extend(nums)`. However this modifies the original list, so it's better to create a copy first.

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

Trade-offs: Slightly more concise in Python but potentially less clear for beginners; creating a copy first is generally preferred for immutability.

List Comprehension (Python)

A Pythonic approach using list comprehension: `ans = nums + nums`

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

Trade-offs: More concise, but may be less readable for beginners than the explicit loop.

✨

Key Insights

1Insight 1: The problem's core is simple array manipulation. No complex data structures or algorithms are needed.
2Insight 2: Directly creating a new array of double the size and copying elements is a straightforward and efficient approach.
3Insight 3: The problem highlights the importance of understanding array indexing and iterating through arrays.

📚

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: An empty input array (`nums` with length 0). The solution handles this correctly by creating an empty `ans` array.
Edge case 2: An array with only one element. The solution correctly duplicates the single element in the `ans` array.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly handling array indices, leading to out-of-bounds errors or incorrect copying. Pay close attention to `i` and `i + n` when accessing `ans`.

✗Common mistake 2: Forgetting to allocate sufficient memory for `ans` (especially in C). In C, you must allocate `2 * n` elements, not just `n`.

✅

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 duplication or mirroring. In some systems, data might need to be stored in duplicate for redundancy or backup purposes. This algorithm can efficiently create those duplicates.

🏦Real-world scenario 2: Signal processing. In signal processing, you might want to concatenate a signal with itself to analyze specific patterns or properties of the repeated signal.

💡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

Even for easy problems, discuss time and space complexity. Interviewers want to see you think analytically.

⚠️ 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: October 8, 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

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 14

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

@DevExpert92Jul 7

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

@AlgoMasterJun 30

❓

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.

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 "Concatenation of Array"?

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

🤖 Need help with this problem?