🔥Easy Solution in C++ || Detailed explanation🔥
Explanation\n\n- In the getConcatenation function, we receive a reference to a vector nums containing integers.\n\n- We find the size of the input array nums an
# Explanation\n\n- In the getConcatenation function, we receive a reference to a vector `nums` containing integers.\n\n- We find the size of the input array `nums` and store it in the variable `n`.\n\n- We initialize a new vector `ans` of size `2 * n` to hold the concatenated array.\n\n- We iterate through the elements of `nums` using a loop from `0` to `n - 1`.\n\n- Inside the loop, for each index `i`, we copy the element at index `i` from `nums` to index `i` in `ans` (the first half of the concatenated array), and also to index `i + n` in `ans` (the second half of the concatenated array).\n\n- Finally, we return the concatenated array `ans`.\n\nThis solution has a time complexity of $$O(n)$$ where `n` is the size of the input array `nums`, as we iterate through the elements of `nums` once to construct the concatenated array.\n\n---\n\n\n\n# Code\n```\nclass Solution {\npublic:\n    vector<int> getConcatenation(vector<int>& nums) {\n        int n = nums.size();\n        vector<int> ans(2 * n);\n\n        for (int i = 0; i < n; ++i) {\n            ans[i] = nums[i];\n            ans[i + n] = nums[i];\n        }\n\n        return ans;\n    }\n};\n\n```\n\n\n![meme.jpg](https://assets.leetcode.com/users/images/8cab591c-dee9-4eea-bb8f-32278894f8cb_1712098290.7045112.jpeg)\n
['Array', 'Simulation', 'C++']