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
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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;Â
// Function to concatenate the// given array arr[] twicevoid concatTwice(int* arr, int N){    // Stores array after    // concatenation    int newArr[2 * N];Â
    // Loop to iterate arr[]    for (int i = 0; i < N; i++) {        newArr[i] = arr[i];        newArr[i + N] = arr[i];    }Â
    // Print Answer    for (int i = 0; i < 2 * N; i++) {        cout << newArr[i] << " ";    }}Â
// Driver Codeint main(){Â Â Â Â int arr[] = { 1, 2, 3 };Â Â Â Â int N = sizeof(arr) / sizeof(arr[0]);Â
    concatTwice(arr, N);Â
    return 0;} |
Java
// Java program for the above approachimport java.io.*;import java.lang.*;import java.util.*;Â
class GFG {Â
  // 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];Â
    // Loop to iterate arr[]    for (int i = 0; i < N; i++) {      newArr[i] = arr[i];      newArr[i + N] = arr[i];    }Â
    // Print Answer    for (int i = 0; i < 2 * N; i++) {      System.out.print(newArr[i] + " ");    }  }Â
  // Driver Code  public static void main (String[] args) {    int arr[] = { 1, 2, 3 };    int N = arr.length;Â
    concatTwice(arr, N);  }}Â
// This code is contributed by hrithikgarg03188 |
Python3
# Python program of the above approachÂ
# Function to concatenate the# given array arr[] twicedef concatTwice(arr, N):Â
    # Stores array after    # concatenation    newArr = [0] * (2 * N)Â
    # Loop to iterate arr[]    for i in range(N):        newArr[i] = arr[i]        newArr[i + N] = arr[i]Â
    # Print Answer    for i in range(0, 2 * N):        print(newArr[i], end=" ")Â
# Driver Codearr = [1, 2, 3]N = len(arr)Â
concatTwice(arr, N)Â
# This code is contributed by gfgking |
C#
// C# program for the above approachusing System;class GFG {Â
  // 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];Â
    // Loop to iterate arr[]    for (int i = 0; i < N; i++) {      newArr[i] = arr[i];      newArr[i + N] = arr[i];    }Â
    // Print Answer    for (int i = 0; i < 2 * N; i++) {      Console.Write(newArr[i] + " ");    }  }Â
  // Driver Code  public static void Main () {    int []arr = { 1, 2, 3 };    int N = arr.Length;Â
    concatTwice(arr, N);  }}Â
// This code is contributed by Samim Hossain Mondal |
Javascript
<script>Â Â Â Â // JavaScript program of the above approachÂ
    // Function to concatenate the    // given array arr[] twice    const concatTwice = (arr, N) => {             // Stores array after        // concatenation        let newArr = new Array(2 * N).fill(0);Â
        // Loop to iterate arr[]        for (let i = 0; i < N; i++) {            newArr[i] = arr[i];            newArr[i + N] = arr[i];        }Â
        // Print Answer        for (let i = 0; i < 2 * N; i++) {            document.write(`${newArr[i]} `);        }    }Â
    // Driver Code    let arr = [1, 2, 3];    let N = arr.length;Â
    concatTwice(arr, N);Â
// This code is contributed by rakeshsahniÂ
</script> |
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Output
1 2 3 1 2 3
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Time Complexity: O(N)
Auxiliary Space: O(N)
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