Array obtained by repeatedly reversing array after every insertion from given array

0 0 5 minutes read

Given an array arr[], the task is to print the array obtained by inserting elements of arr[] one by one into an initially empty array, say arr1[], and reversing the array arr1[] after every insertion.

Examples:

Input: arr[] = {1, 2, 3, 4}
Output: 4 2 1 3
Explanation:
Operations performed on the array arr1[] as follows:
Step 1: Append 1 into the array and reverse it. arr1[] = {1}
Step 2: Append 2 into the array and reverse it. arr1[] = {2, 1}
Step 3: Append 3 into the array and reverse it. arr1[] = {3, 1, 2}
Step 3: Append 4 into the array and reverse it. arr1[] = {4, 2, 1, 3}

Input: arr[] = {1, 2, 3}
Output: 3 1 2
Explanation:
Operations performed on the array arr1[] as follows:
Step 1: Append 1 into the array and reverse it. arr1[] = {1}
Step 2: Append 2 into the array and reverse it. arr1[] = {2, 1}
Step 3: Append 3 into the array and reverse it. arr1[] = {3, 1, 2}

Naive Approach: The simplest approach to solve the problem is to iterate over the array arr[] and insert each element of arr[] one by one into the array arr1[] and reverse the array arr1[] after each insertion.

Time Complexity: O(N²)
Auxiliary Space: O(1)

Efficient Approach: To optimize the above approach, the idea is to use the Doubly Ended Queue (Deque) to append the elements at both ends. Follow the steps below to solve the problem:

Initialize a Double Ended Queue for storing the elements of the converted array.
Iterate over the elements of the array arr[] and push the elements to the front of the Deque if the index of the element and length of the element have the same parity. Otherwise, at the push it to the back of the Deque.
Finally, after the complete iteration of the array arr[], print the contents of the Deque as the required arrangement of output.

Below is the implementation of the above approach:

C++

// C++ program of the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to generate the array by
// inserting array elements one by one
// followed by reversing the array
void generateArray(int arr[], int n)
{
 
    // Doubly ended Queue
    deque<int> ans;
 
    // Iterate over the array
    for (int i = 0; i < n; i++) {
 
        // Push array elements
        // alternately to the front
        // and back
        if (i & 1)
            ans.push_front(arr[i]);
        else
            ans.push_back(arr[i]);
    }
 
    // If size of list is odd
    if (n & 1) {
 
        // Reverse the list
        reverse(ans.begin(),
                ans.end());
    }
 
    // Print the elements
    // of the array
    for (auto x : ans) {
        cout << x << " ";
    }
    cout << endl;
}
 
// Driver Code
int32_t main()
{
    int n = 4;
    int arr[n] = { 1, 2, 3, 4 };
    generateArray(arr, n);
    return 0;
}

Java

// Java program of the above approach 
import java.io.*;
import java.util.*;
 
class GFG{
     
// Function to generate the array by
// inserting array elements one by one
// followed by reversing the array
static void generateArray(int arr[], int n)
{
     
    // Doubly ended Queue
    Deque<Integer> ans = new LinkedList<>();
  
    // Iterate over the array
    for(int i = 0; i < n; i++) 
    {
         
        // Push array elements
        // alternately to the front
        // and back
        if ((i & 1) != 0)
            ans.addFirst(arr[i]);
        else
            ans.add(arr[i]);
    }
  
    // If size of list is odd
    if ((n & 1) != 0) 
    {
         
        // Reverse the list
        Collections.reverse(Arrays.asList(ans)); 
    }
  
    // Print the elements
    // of the array
    for(int x : ans)
    {
        System.out.print(x + " ");
    }
    System.out.println();
}
  
// Driver Code
public static void main (String[] args)
{
    int n = 4;
    int arr[] = { 1, 2, 3, 4 };
     
    generateArray(arr, n);
}
}
 
// This code is contributed by code_hunt

Python3

# Python3 program of the above approach 
from collections import deque
 
# Function to generate the array by
# inserting array elements one by one
# followed by reversing the array
def generateArray(arr, n):
  
    # Doubly ended Queue
    ans = deque()
  
    # Iterate over the array
    for i in range(n):
  
        # Push array elements
        # alternately to the front
        # and back
        if (i & 1 != 0):
            ans.appendleft(arr[i])
        else:
            ans.append(arr[i])
     
    # If size of list is odd
    if (n & 1 != 0):
  
        # Reverse the list
        ans.reverse()
     
    # Print the elements
    # of the array
    for x in ans:
        print(x, end = " ")
     
    print()
 
# Driver Code
n = 4
arr = [ 1, 2, 3, 4 ]
 
generateArray(arr, n)
 
# This code is contributed by code_hunt

C#

// C# program of 
// the above approach 
using System;
using System.Collections.Generic;
class GFG{
     
// Function to generate the array 
// by inserting array elements 
// one by one followed by 
// reversing the array
static void generateArray(int []arr, 
                          int n)
{    
  // Doubly ended Queue
  List<int> ans = new List<int>();
 
  // Iterate over the array
  for(int i = 0; i < n; i++) 
  {
    // Push array elements
    // alternately to the front
    // and back
    if ((i & 1) != 0)
      ans.Insert(0, arr[i]);
    else
      ans.Add(arr[i]);
  }
 
  // If size of list is odd
  if ((n & 1) != 0) 
  {
    // Reverse the list
    ans.Reverse(); 
  }
 
  // Print the elements
  // of the array
  foreach(int x in ans)
  {
    Console.Write(x + " ");
  }
  Console.WriteLine();
}
  
// Driver Code
public static void Main(String[] args)
{
  int n = 4;
  int []arr = {1, 2, 3, 4};
  generateArray(arr, n);
}
}
 
// This code is contributed by 29AjayKumar

Javascript

<script>
 
// Javascript program of the above approach
 
// Function to generate the array by
// inserting array elements one by one
// followed by reversing the array
function generateArray(arr, n)
{
 
    // Doubly ended Queue
    var ans = [];
 
    // Iterate over the array
    for (var i = 0; i < n; i++) {
 
        // Push array elements
        // alternately to the front
        // and back
        if (i & 1)
            ans.splice(0,0,arr[i]);
        else
            ans.push(arr[i]);
    }
 
    // If size of list is odd
    if (n & 1) {
 
        // Reverse the list
        ans.reverse();
    }
 
    // Print the elements
    // of the array
    ans.forEach(x => {
 
        document.write( x + " ");
    });
}
 
// Driver Code
var n = 4;
var arr =  [1, 2, 3, 4];
generateArray(arr, n);
 
</script>

Output:

4 2 1 3

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

Recommended

Solve DSA problems on GfG Practice.

Solve Problems

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!