Concatenation of Array

In this tutorial, we will solve the Concatenate Array problem using a simple approach and an optimized approach. We will provide the implementation of the solution in C++, Java, and Python.

Problem Description

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 array ans.

Examples

Example 1:

Input: nums = [1,2,1]
Output: [1,2,1,1,2,1]

Example 2:

Input: nums = [1,3,2,1]
Output: [1,3,2,1,1,3,2,1]

Constraints

• n == nums.length
• 1 <= n <= 1000

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Solution for Concatenate Array Problem

Approach 1: Simple Concatenation

The approach is straightforward: iterate through the array nums and copy each element twice, once to the first half and once to the second half of the result array ans.

C++

Written by @ImmidiSivani

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;
    }
};

Java

Written by @ImmidiSivani

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;
    }
}

Python

Written by @ImmidiSivani

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

Approach 2: Optimized Concatenation

An optimized approach for concatenation in some languages can take advantage of built-in functions to concatenate the array directly.

C++

Written by @ImmidiSivani

class Solution {
public:
    vector<int> getConcatenation(vector<int>& nums) {
        vector<int> ans = nums;
        ans.insert(ans.end(), nums.begin(), nums.end());
        return ans;
    }
};

Java

Written by @ImmidiSivani

import java.util.*;

class Solution {
    public int[] getConcatenation(int[] nums) {
        int n = nums.length;
        int[] ans = Arrays.copyOf(nums, 2 * n);
        System.arraycopy(nums, 0, ans, n, n);
        return ans;
    }
}

Python

Written by @ImmidiSivani

class Solution:
    def getConcatenation(self, nums: List[int]) -> List[int]:
        return nums + nums

Complexity Analysis

• Time Complexity: $O(n)$, where n is the length of the input array nums. We iterate through the array once.
• Space Complexity: $O(n)$, since we are using a new array ans of size 2n.

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Authors:

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