In this blog post, we’ll explore a coding problem from LeetCode that involves concatenating an array with itself. We’ll discuss the problem statement and provide a solution using a simple algorithm.

The problem is as follows:

Given an integer array `nums`

of length `n`

, the task is to create a new array `ans`

of length `2n`

such that `ans[i] == nums[i]`

and `ans[i + n] == nums[i]`

for `0 <= i < n`

(0-indexed). In simpler terms, the array `ans`

is formed by concatenating two instances of the original array `nums`

.

To solve this problem, we can use a straightforward approach. We create a new array `ans`

with a length of `2n`

and fill it by copying elements from the original array `nums`

twice. Here’s the solution in C#:

```
public class Solution {
public int[] GetConcatenation(int[] nums) {
int length = nums.Length;
int[] newArray = new int[2 * length];
for (int i = 0; i < length; i++)
newArray[i] = newArray[i + length] = nums[i];
return newArray;
}
}
```

In this solution, we iterate through the original array `nums`

and copy each element to the corresponding positions in the new array `newArray`

. The result is a concatenated array that meets the given conditions.

### Conclusion

This LeetCode problem provides a simple exercise in array manipulation and indexing. The solution presented here demonstrates a basic algorithmic approach to solving the problem. Understanding and practicing such problems are valuable for improving algorithmic thinking and problem-solving skills in programming.

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