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 &lt;= i &lt; n (0-indexed).

Specifically, ans is the concatenation of two nums arrays.

Return the array ans.

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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]


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Constraints:


	n == nums.length
	1 &lt;= n &lt;= 1000
	1 &lt;= nums[i] &lt;= 1000


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

        return ans;
    }
}
class Solution {
public:
    vector<int> getConcatenation(vector<int>& nums) {
        int n=nums.size();
        for(int i=0;i<n;i++)
        {
            nums.push_back(nums[i]);
        }
        return nums;
    }
};
var getConcatenation = function(nums) {
	//spread the nums array twice and return it
    return [...nums,...nums]
};
Concatenation of Array