# 1929. Concatenation of Array

Today, we’ll be looking at the following problem:

1929. Concatenation of Array

# Prompt

We are given an array nums that has a total length of n elements. We want to create an array called ans that has a total length of 2n elements. The problem states that ans[i] == nums[i] and ans[i + n] == nums[i] meaning that at whichever index, the value located at that index will be identical to the value located at that index + n.

If this still isn’t clear, let’s observe the 1st example they provide. You can see that [1, 2, 1] is given as the input and [1, 2, 1, 1, 2, 1] is the output. If we take the value 2 as an example from the original array, the value 2 is located at index 1. In this case, if we add 1 with the length of the original array n, then the output array or ans at index 1 + 3 will also have a value of 2.

Input: [1, 2, 1]
Index: 0, 1, 2

Output: [1, 2, 1, 1, 2, 1]
Index: 0, 1, 2, 3, 4, 5

# Pseudocode

Unless it’s specifically stated in the prompt, you either have the choice of appending each element back into the array as you iterate through it in-place or you can instantiate a new array and populate that array with 2n elements. My code will reflect the former as I wanted to save space by performing the algorithm in-place.

# Code

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## More from John Kim

iOS Developer | Full Stack Developer | Software Engineer | LinkedIn: john-kim-developer | GitHub: cloudiosx | Portfolio: cloudiosx.com

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iOS Developer | Full Stack Developer | Software Engineer | LinkedIn: john-kim-developer | GitHub: cloudiosx | Portfolio: cloudiosx.com