Concatenate given array twice

Given an array arr[] of N elements, the task is to concatenate it twice, i.e. create an array of size 2*N by appending the copy of the given array to itself.

Example:

Input: arr[] = {1, 2, 1}
Output: 1 2 1 1 2 1
Explanation: The given array arr[] = {1, 2, 1}, can be appended to itself resulting in arr[] = {1, 2, 1, 1, 2, 1}.

Input: arr[] = {1, 3, 2, 1}
Output: 1 3 2 1 1 3 2 1

Approach: The given problem is an implementation-based problem. It can be solved by creating a array newArr[] of size 2*N. Iterate the given array arr[] using a variable i in the range [0, N) and append the assign newArr[i] = arr[i] and newArr[i + N] = arr[i].

Below is the implementation of the above approach:

C++

// C++ program of the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to concatenate the
// given array arr[] twice
void concatTwice(int* arr, int N)
{
    // Stores array after
    // concatenation
    int newArr[2 * N];

    // Loop to iterate arr[]
    for (int i = 0; i < N; i++) {
        newArr[i] = arr[i];
        newArr[i + N] = arr[i];
    }

    // Print Answer
    for (int i = 0; i < 2 * N; i++) {
        cout << newArr[i] << " ";
    }
}

// Driver Code
int main()
{
    int arr[] = { 1, 2, 3 };
    int N = sizeof(arr) / sizeof(arr[0]);

    concatTwice(arr, N);

    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;

class GFG {

  // Function to concatenate the
  // given array arr[] twice
  static void concatTwice(int arr[], int N)
  {
    // Stores array after
    // concatenation
    int newArr[] = new int[2 * N];

    // Loop to iterate arr[]
    for (int i = 0; i < N; i++) {
      newArr[i] = arr[i];
      newArr[i + N] = arr[i];
    }

    // Print Answer
    for (int i = 0; i < 2 * N; i++) {
      System.out.print(newArr[i] + " ");
    }
  }

  // Driver Code
  public static void main (String[] args) {
    int arr[] = { 1, 2, 3 };
    int N = arr.length;

    concatTwice(arr, N);
  }
}

// This code is contributed by hrithikgarg03188

Python3

# Python program of the above approach

# Function to concatenate the
# given array arr[] twice
def concatTwice(arr, N):

    # Stores array after
    # concatenation
    newArr = [0] * (2 * N)

    # Loop to iterate arr[]
    for i in range(N):
        newArr[i] = arr[i]
        newArr[i + N] = arr[i]

    # Print Answer
    for i in range(0, 2 * N):
        print(newArr[i], end=" ")

# Driver Code
arr = [1, 2, 3]
N = len(arr)

concatTwice(arr, N)

# This code is contributed by gfgking

C#

// C# program for the above approach
using System;
class GFG {

  // Function to concatenate the
  // given array arr[] twice
  static void concatTwice(int []arr, int N)
  {

    // Stores array after
    // concatenation
    int []newArr = new int[2 * N];

    // Loop to iterate arr[]
    for (int i = 0; i < N; i++) {
      newArr[i] = arr[i];
      newArr[i + N] = arr[i];
    }

    // Print Answer
    for (int i = 0; i < 2 * N; i++) {
      Console.Write(newArr[i] + " ");
    }
  }

  // Driver Code
  public static void Main () {
    int []arr = { 1, 2, 3 };
    int N = arr.Length;

    concatTwice(arr, N);
  }
}

// This code is contributed by Samim Hossain Mondal

Javascript

<script>
    // JavaScript program of the above approach

    // Function to concatenate the
    // given array arr[] twice
    const concatTwice = (arr, N) => {

        // Stores array after
        // concatenation
        let newArr = new Array(2 * N).fill(0);

        // Loop to iterate arr[]
        for (let i = 0; i < N; i++) {
            newArr[i] = arr[i];
            newArr[i + N] = arr[i];
        }

        // Print Answer
        for (let i = 0; i < 2 * N; i++) {
            document.write(`${newArr[i]} `);
        }
    }

    // Driver Code
    let arr = [1, 2, 3];
    let N = arr.length;

    concatTwice(arr, N);

// This code is contributed by rakeshsahni

</script>

Output

1 2 3 1 2 3 

Time Complexity: O(N)
Auxiliary Space: O(N)