Concatenation of Array easy

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. 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]

Steps

1. Create the result array: Initialize an array ans with a length twice that of the input array nums.
2. Copy elements: Iterate through the input array nums. For each element at index i, copy it to ans[i] and ans[i + nums.Length].

Explanation

The problem requires creating a new array that contains the elements of the input array twice. The solution directly implements this by creating a larger array and copying the elements to the appropriate positions. The key is understanding that the second half of the new array is a direct copy of the first half.

Code

public class Solution {
    public int[] GetConcatenation(int[] nums) {
        int n = nums.Length;
        int[] ans = new int[2 * n];

        for (int i = 0; i < n; i++) {
            ans[i] = nums[i];
            ans[i + n] = nums[i];
        }

        return ans;
    }
}

Complexity

• Time Complexity: O(n), where n is the length of the input array. We iterate through the input array once.
• Space Complexity: O(n), because we create a new array of size 2n to store the result. Note that the space used by the input array is not considered in the space complexity analysis.