Problem

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 (0-indexed).

Specifically, ans is the concatenation of two nums arrays.

Return the arrayans.

Examples

Example 1

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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

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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]

Constraints

  • n == nums.length
  • 1 <= n <= 1000
  • 1 <= nums[i] <= 1000

Solution

Method 1 – Simple Array Concatenation

Intuition

The task is to create a new array by repeating the input array twice in order. This is a direct simulation problem with no tricky edge cases.

Approach

  1. Let n be the length of nums.
  2. Create a new array ans of length 2n.
  3. For each i from 0 to n-1:
    • Set ans[i] = nums[i]
    • Set ans[i + n] = nums[i]
  4. Return ans.

Code

C++
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class Solution {
public:
    vector<int> getConcatenation(vector<int>& nums) {
        int n = nums.size();
        vector<int> ans(2 * n);
        for (int i = 0; i < n; ++i) {
            ans[i] = nums[i];
            ans[i + n] = nums[i];
        }
        return ans;
    }
};
Go
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func GetConcatenation(nums []int) []int {
    n := len(nums)
    ans := make([]int, 2*n)
    for i := 0; i < n; i++ {
        ans[i] = nums[i]
        ans[i+n] = nums[i]
    }
    return ans
}
Java
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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;
    }
}
Kotlin
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class Solution {
    fun getConcatenation(nums: IntArray): IntArray {
        val n = nums.size
        val ans = IntArray(2 * n)
        for (i in 0 until n) {
            ans[i] = nums[i]
            ans[i + n] = nums[i]
        }
        return ans
    }
}
Python
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class Solution:
    def getConcatenation(self, nums: list[int]) -> list[int]:
        n = len(nums)
        ans = [0] * (2 * n)
        for i in range(n):
            ans[i] = nums[i]
            ans[i + n] = nums[i]
        return ans
Rust
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impl Solution {
    pub fn get_concatenation(nums: Vec<i32>) -> Vec<i32> {
        let n = nums.len();
        let mut ans = vec![0; 2 * n];
        for i in 0..n {
            ans[i] = nums[i];
            ans[i + n] = nums[i];
        }
        ans
    }
}
TypeScript
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class Solution {
    getConcatenation(nums: number[]): number[] {
        const n = nums.length;
        const ans = new Array(2 * n);
        for (let 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 nums. We copy each element twice.
  • 🧺 Space complexity: O(n), for the output array of size 2n.