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[] twicevoid 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 Codeint main(){ int arr[] = { 1, 2, 3 }; int N = sizeof(arr) / sizeof(arr[0]); concatTwice(arr, N); return 0;} |
Java
// Java program for the above approachimport 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[] twicedef 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 Codearr = [1, 2, 3]N = len(arr)concatTwice(arr, N)# This code is contributed by gfgking |
C#
// C# program for the above approachusing 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)
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