---
description: 'Author: @heiheihang | https://leetcode.com/problems/concatenation-of-array/'
---

# 1929 - Concatenation of Array (Easy)

## Problem Link

https://leetcode.com/problems/concatenation-of-array/

## 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]
Explanation: The array ans is formed as follows:
- ans = [nums[0],nums[1],nums[2],nums[3],nums[0],nums[1],nums[2],nums[3]]
- ans = [1,3,2,1,1,3,2,1]
```

**Constraints:**

* `n == nums.length`
* `1 <= n <= 1000`
* `1 <= nums[i] <= 1000`

## Approach 1: Iteration

We need to look at `nums` two times to create the desired result. The simplest approach is to perform two for-loops in `nums` and copy the numbers to `result` .

<SolutionAuthor name="@heiheihang"/>

```python
def getConcatenation(self, nums: List[int]) -> List[int]:
        
        #initialize result
        result = []
        
        #first iteration of nums
        for i in range(len(nums)):
            result.append(nums[i])
        
        #second iteration of nums
        for i in range(len(nums)):
            result.append(nums[i])
        
        #return result
        return result
```
