Find the final sequence of the array after performing given operations

Given an array arr[] of size N, the task is to perform the following operation exactly N times,
Create an empty list of integers b[] and in the ith operation,

  1. Append arr[i] to the end of b[].
  2. Reverse the elements in b[].

Finally, print the contents of the list b[] after the end of all operations.

Examples:

Input: arr[] = {1, 2, 3, 4}
Output: 4 2 1 3

Operation b[]
1 {1}
2 {2, 1}
3 {3, 1, 2}
4 {4, 2, 1, 3}

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

Approach: We need some observations to solve this problem. Suppose the number of elements in the array is even. Say our array is {4, 8, 6, 1, 7, 9}.

Operation b[]
1 {4}
2 {8, 4}
3 {6, 4, 8}
4 {1, 8, 4, 6}
5 {7, 6, 4, 8, 1}
6 {9, 1, 8, 4, 6, 7}

After carefully observing, we conclude that for even size of elements in the array the numbers which are at even positions (index 1 based) are reversed and added at the beginning and the numbers which are at the odd positions are kept in same order and added in the end.

While for odd-sized arrays, the elements at odd positions are reversed and added at the beginning while elements in the array at even positions are kept same and added in the end.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function that generates the
// array b[] when n is even
void solveEven(int n, int* arr, int* b)
{
    int left = n - 1;
  
    // Fill the first half of the final array
    // with reversed sequence
    for (int i = 0; i < (n / 2); ++i) {
        b[i] = arr[left];
        left = left - 2;
        if (left < 0)
            break;
    }
  
    // Fill the second half
    int right = 0;
    for (int i = n / 2; i <= n - 1; ++i) {
        b[i] = arr[right];
        right = right + 2;
        if (right > n - 2)
            break;
    }
}
  
// Function that generates the
// array b[] when n is odd
void solveOdd(int n, int* arr, int* b)
{
  
    // Fill the first half of the final array
    // with reversed sequence
    int left = n - 1;
    for (int i = 0; i < (n / 2) + 1; ++i) {
        b[i] = arr[left];
        left = left - 2;
        if (left < 0)
            break;
    }
  
    // Fill the second half
    int right = 1;
    for (int i = (n / 2) + 1; i <= n - 1; ++i) {
        b[i] = arr[right];
        right = right + 2;
        if (right > n - 2)
            break;
    }
}
  
// Function to find the final array b[]
// after n operations of given type
void solve(int n, int* arr)
{
  
    // Create the array b
    int b[n];
  
    // If the array size is even
    if (n % 2 == 0)
        solveEven(n, arr, b);
    else
        solveOdd(n, arr, b);
  
    // Print the final array elements
    for (int i = 0; i <= n - 1; ++i) {
        cout << b[i] << " ";
    }
}
  
// Driver code
int main()
{
    int arr[] = { 1, 2, 3, 4 };
    int n = sizeof(arr) / sizeof(arr[0]);
  
    solve(n, arr);
  
    return 0;
}

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Java

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// Java implementation of the approach
import java.io.*;
  
class GFG 
{
  
// Function that generates the
// array b[] when n is even
static void solveEven(int n, int arr[], int b[])
{
    int left = n - 1;
  
    // Fill the first half of the final array
    // with reversed sequence
    for (int i = 0; i < (n / 2); ++i)
    {
        b[i] = arr[left];
        left = left - 2;
        if (left < 0)
            break;
    }
  
    // Fill the second half
    int right = 0;
    for (int i = n / 2; i <= n - 1; ++i)
    {
        b[i] = arr[right];
        right = right + 2;
        if (right > n - 2)
            break;
    }
}
  
// Function that generates the
// array b[] when n is odd
static void solveOdd(int n, int arr[], int b[])
{
  
    // Fill the first half of the final array
    // with reversed sequence
    int left = n - 1;
    for (int i = 0; i < (n / 2) + 1; ++i) 
    {
        b[i] = arr[left];
        left = left - 2;
        if (left < 0)
            break;
    }
  
    // Fill the second half
    int right = 1;
    for (int i = (n / 2) + 1; i <= n - 1; ++i) 
    {
        b[i] = arr[right];
        right = right + 2;
        if (right > n - 2)
            break;
    }
}
  
// Function to find the final array b[]
// after n operations of given type
static void solve(int n, int arr[])
{
  
    // Create the array b
    int b[] = new int[n];
  
    // If the array size is even
    if (n % 2 == 0)
        solveEven(n, arr, b);
    else
        solveOdd(n, arr, b);
  
    // Print the final array elements
    for (int i = 0; i <= n - 1; ++i)
    {
        System.out.print( b[i] + " ");
    }
}
  
// Driver code
public static void main (String[] args) 
{
    int []arr = { 1, 2, 3, 4 };
    int n = arr.length;
      
    solve(n, arr);
}
}
  
// This code is contributed by anuj_67..

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Python3

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# Python 3 implementation of the approach
  
# Function that generates the
# array b[] when n is even
def solveEven(n, arr, b):
    left = n - 1
  
    # Fill the first half of the final array
    # with reversed sequence
    for i in range((n // 2)):
        b[i] = arr[left]
        left = left - 2
        if (left < 0):
            break
  
    # Fill the second half
    right = 0
    for i in range(n // 2, n, 1):
        b[i] = arr[right]
        right = right + 2
        if (right > n - 2):
            break
  
# Function that generates the
# array b[] when n is odd
def solveOdd(n, arr, b):
      
    # Fill the first half of the final array
    # with reversed sequence
    left = n - 1
    for i in range(n // 2 + 1):
        b[i] = arr[left]
        left = left - 2
        if (left < 0):
            break
  
    # Fill the second half
    right = 1
    for i in range(n // 2 + 1, n, 1):
        b[i] = arr[right]
        right = right + 2
        if (right > n - 2):
            break
  
# Function to find the final array b[]
# after n operations of given type
def solve(n, arr):
      
    # Create the array b
    b = [0 for i in range(n)]
  
    # If the array size is even
    if (n % 2 == 0):
        solveEven(n, arr, b)
    else:
        solveOdd(n, arr, b)
  
    # Print the final array elements
    for i in range(n):
        print(b[i], end = " ")
  
# Driver code
if __name__ == '__main__':
    arr = [1, 2, 3, 4]
    n = len(arr)
  
    solve(n, arr)
  
# This code is contributed by
# Surendra_Gangwar

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C#

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// C# implementation of the approach
using System;
  
class GFG 
{
  
// Function that generates the
// array b[] when n is even
static void solveEven(int n, int []arr, 
                             int []b)
{
    int left = n - 1;
  
    // Fill the first half of the final array
    // with reversed sequence
    for (int i = 0; i < (n / 2); ++i)
    {
        b[i] = arr[left];
        left = left - 2;
        if (left < 0)
            break;
    }
  
    // Fill the second half
    int right = 0;
    for (int i = n / 2; i <= n - 1; ++i)
    {
        b[i] = arr[right];
        right = right + 2;
        if (right > n - 2)
            break;
    }
}
  
// Function that generates the
// array b[] when n is odd
static void solveOdd(int n, int []arr, int []b)
{
  
    // Fill the first half of the final array
    // with reversed sequence
    int left = n - 1;
    for (int i = 0; i < (n / 2) + 1; ++i) 
    {
        b[i] = arr[left];
        left = left - 2;
        if (left < 0)
            break;
    }
  
    // Fill the second half
    int right = 1;
    for (int i = (n / 2) + 1; i <= n - 1; ++i) 
    {
        b[i] = arr[right];
        right = right + 2;
        if (right > n - 2)
            break;
    }
}
  
// Function to find the final array b[]
// after n operations of given type
static void solve(int n, int []arr)
{
  
    // Create the array b
    int []b = new int[n];
  
    // If the array size is even
    if (n % 2 == 0)
        solveEven(n, arr, b);
    else
        solveOdd(n, arr, b);
  
    // Print the final array elements
    for (int i = 0; i <= n - 1; ++i)
    {
        Console.Write( b[i] + " ");
    }
}
  
// Driver code
public static void Main () 
{
    int []arr = { 1, 2, 3, 4 };
    int n = arr.Length;
      
    solve(n, arr);
}
}
  
// This code is contributed by anuj_67..

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Output:

4 2 1 3

Efficient Approach:
The last Element of Array will be reversed only once. Last but one element will be reversed twice. Hence it goes to the last position in final result array ie b. Hence we can fill b array by iterating the original array from the end and placing elements at the not filled first index and not filled the last index. The same idea is implemented below.

Below is the implementation of the above approach:

Java

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// Java implementation of the approach
import java.util.*;
import java.lang.*;
import java.io.*;
  
class GFG {
  
    static int[] solve(int[] arr, int n) {
        int[] b = new int[n];
        int p = 0;
        for (int i = n - 1; i >= 0; i--) {
            b[p] = arr[i--];
            if (i >= 0)
                b[n - 1 - p] = arr[i];
            p++;
        }
        return b;
    }
  
    public static void main(String[] args)
    {
        int []arr = { 1, 2, 3, 4 };
        int n = arr.length;
          
        int[] b = solve(arr, n);
          
        System.out.println(Arrays.toString(b));
    }
}
  
// This code is contributed by Pramod Hosahalli

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C#

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// C# implementation of the approach
using System;
      
class GFG 
{
    static int[] solve(int[] arr, int n) 
    {
        int[] b = new int[n];
        int p = 0;
        for (int i = n - 1; i >= 0; i--) 
        {
            b[p] = arr[i--];
            if (i >= 0)
                b[n - 1 - p] = arr[i];
            p++;
        }
        return b;
    }
  
    // Driver Code
    public static void Main(String[] args)
    {
        int []arr = { 1, 2, 3, 4 };
        int n = arr.Length;
          
        int[] b = solve(arr, n);
          
        Console.WriteLine("[" + String.Join(",", b) + "]");
    }
}
  
// This code is contributed by Princi Singh

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Output:

[4, 2, 1, 3]


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