Concatenation of Array

Difficulty: Easy Source: NeetCode

Problem

Given an integer array nums of length n, return an array ans of length 2n where ans[i] = nums[i] and ans[i + n] = nums[i] for 0 <= i < n.

In other words, ans is the concatenation of two copies of nums.

Example 1: Input: nums = [1, 2, 1] Output: [1, 2, 1, 1, 2, 1]

Example 2: Input: nums = [1, 3, 2, 1] Output: [1, 3, 2, 1, 1, 3, 2, 1]

Constraints:

• 1 <= n <= 1000
• 1 <= nums[i] <= 1000

Prerequisites

Before attempting this problem, you should be comfortable with:

• Arrays — understanding index-based access and iteration
• Array concatenation — combining arrays or building new ones by index

1. Brute Force

Intuition

Build the result array manually by iterating twice. On the first pass, place each element of nums at index i. On the second pass, place the same elements at index i + n. This makes the logic explicit and easy to follow, even if Python gives us a nicer shortcut.

Algorithm

1. Create ans as a list of zeros with length 2 * n.
2. Loop i from 0 to n - 1:
   • Set ans[i] = nums[i]
   • Set ans[i + n] = nums[i]
3. Return ans.

Solution

def getConcatenation(nums):
    n = len(nums)
    ans = [0] * (2 * n)
    for i in range(n):
        ans[i] = nums[i]
        ans[i + n] = nums[i]
    return ans


print(getConcatenation([1, 2, 1]))        # [1, 2, 1, 1, 2, 1]
print(getConcatenation([1, 3, 2, 1]))     # [1, 3, 2, 1, 1, 3, 2, 1]
print(getConcatenation([5]))              # [5, 5]

Complexity

• Time: O(n)
• Space: O(n) — the output array of length 2n

2. Python Concatenation Operator

Intuition

Python’s + operator on lists creates a new list that is the concatenation of both operands. Since we want nums followed by nums, we can just return nums + nums — one line, same time and space complexity as the manual approach, just more readable.

This is not a “trick” — it’s the same algorithm underneath. The + operator iterates through both lists and copies each element into a new list, which is exactly what the brute force does.

Algorithm

1. Return nums + nums.

flowchart LR
    A(["nums = [1, 2, 1]"]) --> B["nums + nums"]
    B --> C(["[1, 2, 1, 1, 2, 1]"])

Solution

def getConcatenation(nums):
    return nums + nums


print(getConcatenation([1, 2, 1]))        # [1, 2, 1, 1, 2, 1]
print(getConcatenation([1, 3, 2, 1]))     # [1, 3, 2, 1, 1, 3, 2, 1]
print(getConcatenation([5]))              # [5, 5]

Complexity

• Time: O(n)
• Space: O(n) — output array of length 2n

Common Pitfalls

Using nums * 2 vs nums + nums. Both produce the same result for this problem, but understand that nums * 2 also creates a new list — it does not mutate nums in place. Either works fine here.

Mutating nums in place. If you try to append to nums directly inside a loop over nums, you will get an infinite loop since the list grows as you iterate. Always build a separate output or use nums + nums to create a new list.