Concatenate given array twice
Last Updated :
17 Feb, 2022
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 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
# 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# 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
<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>
Time Complexity: O(N)
Auxiliary Space: O(N)