Skip to content
Related Articles

Related Articles

Improve Article

Rearrange the Array by shifting middle elements to start and end alternatively

  • Difficulty Level : Medium
  • Last Updated : 21 Sep, 2021

Given an array, the task is to shift the middle element to the start and end of the array alternatively, till the middle element becomes equal to the first element of the original array.

Input: arr[]=[2, 8, 5, 9, 10]
Output: [9, 5, 2, 10, 8]
Explanation: We can get this output by shifting middle element
step1: middle element 5 is shifted to front of array [5, 2, 8, 9, 10]
step2: middle element 8 is shifted to end of array [5, 2, 9, 10, 8]
step3: middle element 9 is shifted to front of array [9, 5, 2, 10, 8]

Input: arr[]=[10, 12, 6, 5, 3, 1]
Output: [1, 3, 5, 10, 6, 12]

Naive Approach: Shift the middle element of the array alternatively to the start and end of the array.
Take the middle element and shift it to the start of the Array if c is even or shift it to the end of Array if c is odd. Terminate the loop when the middle element becomes equal to the first element of the Original Array.



Below is the implementation of the above approach :

C++




// C++ program for above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function for shifting middle element.
void AlternateShift(vector<int>& arr, int x)
{
 
    // get middle index
    int mid = arr.size() / 2;
 
    // initialize c to 0
    int c = 0;
   
    // Shift middle element
    // till its value not equals to x.
    while (arr[mid] != x) {
 
        // pop middle element
 
        int z = arr[mid];
        arr.erase(arr.begin() + mid);
 
        // if c is even then insert z
        // at start of the array
        if (c % 2 == 0)
            arr.insert(arr.begin() + 0, z);
 
        // if c is odd then insert z
        // at end of the array
        else
            arr.push_back(z);
 
        // increment count c
        c += 1;
    }
}
 
int main()
{
    vector<int> Arr = { 2, 8, 5, 9, 10 };
 
    // initialize a to zero index array value
    int a = Arr[0];
 
    // call AlternateShift function
    AlternateShift(Arr, a);
 
    // print the changed array Unpacking array
    for (int i = 0; i < Arr.size(); ++i) {
        cout << Arr[i] << " ";
    }
 
    return 0;
}
 
    // This code is contributed by rakeshsahni

Python3




# Function for shifting middle element.
def AlternateShift(arr, x):
 
    # get middle index
    mid = len(arr) // 2
 
    # initialize c to 0
    c = 0
 
    # Shift middle element
    # till its value not equals to x.
    while arr[mid] != x:
 
        # pop middle element
        z = arr.pop(mid)
 
        # if c is even then insert z
        # at start of the array
        if c % 2 == 0:
            arr.insert(0, z)
 
        # if c is odd then insert z
        # at end of the array
        else:
            arr.append(z)
 
        # increment count c
        c += 1
 
 
Arr = [2, 8, 5, 9, 10]
 
# initialize a to zero index array value
a = Arr[0]
 
# call AlternateShift function
AlternateShift(Arr, a)
 
# print the changed array Unpacking array
print(*Arr)

Javascript




<script>
    // JavaScript program for above approach
 
    // Function for shifting middle element.
    const AlternateShift = (arr, x) => {
 
        // get middle index
        let mid = parseInt(arr.length / 2);
 
        // initialize c to 0
        let c = 0;
 
        // Shift middle element
        // till its value not equals to x.
        while (arr[mid] != x) {
 
            // pop middle element
 
            let z = arr[mid];
            arr.splice(mid, 1);
 
            // if c is even then insert z
            // at start of the array
            if (c % 2 == 0)
                arr.splice(0, 0, z);
 
            // if c is odd then insert z
            // at end of the array
            else
                arr.push(z);
 
            // increment count c
            c += 1;
        }
    }
 
    Arr = [2, 8, 5, 9, 10];
 
    // initialize a to zero index array value
    let a = Arr[0];
 
    // call AlternateShift function
    AlternateShift(Arr, a);
 
    // print the changed array Unpacking array
 
    for (let i = 0; i < Arr.length; ++i) {
        document.write(`${Arr[i]} `);
    }
 
// This code is contributed by rakeshsahni
 
</script>
Output
9 5 2 10 8

Time Complexity: O(n2)
Auxiliary Space: O(1)

Efficient Approach: Alternate Shifting is also the case of half reversal of array. First take an element from last to mid if n is even or take an element from last second to mid and insert into new array br[], then insert the first element into br[]. Then insert element from mid-1 to index 1 and insert into br[]. So it will return the array in half reversal order.

Algorithm :

step1: Declare new array br and initialize pos as n-1.
step2: If n is even traverse from last index pos or if n is odd then traverse from second last index pos-1.
step3: Store element from pos index to mid index into br array.
step4: Then insert first element of array to br array. 
step5: If n is odd then insert last value elementof array to br array.
step6: Store element from mid-1 index to index 1 into br array.
step7: return br array.

Below is the implementation of the above algorithm :

C++




// C++ Program of the above approach
#include <iostream>
using namespace std;
 
// Function to to shift the middle
// element to the start and end of
// the array alternatively, till
// the middle element becomes equal to
// the first element of the original Array
int* rearrange(int* ar, int n)
{
    // creating the array to store
    // rearranged value
    int* br = new int[n];
 
    // initialising pos to last index
    int pos = n - 1;
 
    // if n is odd then we will
    // transverse the array
    // from second last element
    if (n % 2 != 0)
        pos = pos - 1;
 
    // storing index of middle element
    int mid = n / 2;
 
    // index variable for rearranged array
    int c = 0;
 
    // transversing the array from
    // the pos to mid index
    // and storing it in br[] array
    for (; pos >= mid; pos--)
        br = ar[pos];
 
    // storing the first element as
    // mid value
    br = ar[0];
 
    // if n is odd then store
    // the last value in br[] the
    // transverse till 1st index
    if (n % 2 != 0)
        br = ar[n - 1];
 
    // storing the first element of
    // array as mid value
    for (; pos >= 1; pos--)
        br = ar[pos];
 
    // returning br[] array
    return br;
}
// Driver Code
int main()
{
    int ar[] = { 2, 8, 5, 9, 10 };
    int n = sizeof(ar) / sizeof(ar[0]);
 
    // Function Call
    int* res = rearrange(ar, n);
 
    // Print answer
    for (int i = 0; i < n; i++)
        cout << res[i] << " ";
}

Java




// Java Program of the above approach
import java.util.*;
 
class GFG
{
 
// Function to to shift the middle
// element to the start and end of
// the array alternatively, till
// the middle element becomes equal to
// the first element of the original Array
static int[] rearrange(int[] ar, int n)
{
   
    // creating the array to store
    // rearranged value
    int[] br = new int[n];
 
    // initialising pos to last index
    int pos = n - 1;
 
    // if n is odd then we will
    // transverse the array
    // from second last element
    if (n % 2 != 0)
        pos = pos - 1;
 
    // storing index of middle element
    int mid = n / 2;
 
    // index variable for rearranged array
    int c = 0;
 
    // transversing the array from
    // the pos to mid index
    // and storing it in br[] array
    for (; pos >= mid; pos--)
        br = ar[pos];
 
    // storing the first element as
    // mid value
    br = ar[0];
 
    // if n is odd then store
    // the last value in br[] the
    // transverse till 1st index
    if (n % 2 != 0)
        br = ar[n - 1];
 
    // storing the first element of
    // array as mid value
    for (; pos >= 1; pos--)
        br = ar[pos];
 
    // returning br[] array
    return br;
}
   
// Driver Code
public static void main(String[] args)
{
    int ar[] = { 2, 8, 5, 9, 10 };
    int n = ar.length;
 
    // Function Call
    int[] res = rearrange(ar, n);
 
    // Print answer
    for (int i = 0; i < n; i++)
        System.out.print(res[i]+ " ");
}
}
 
// This code is contributed by Amit Katiyar

Python3




# Python 3 Program of the above approach
 
# Function to to shift the middle
# element to the start and end of
# the array alternatively, till
# the middle element becomes equal to
# the first element of the original Array
def rearrange(ar, n):
    # creating the array to store
    # rearranged value
    br = [0 for i in range(n)]
 
    # initialising pos to last index
    pos = n - 1
 
    # if n is odd then we will
    # transverse the array
    # from second last element
    if (n % 2 != 0):
        pos = pos - 1
 
    # storing index of middle element
    mid = n // 2
 
    # index variable for rearranged array
    c = 0
 
    # transversing the array from
    # the pos to mid index
    # and storing it in br[] array
    while(pos >= mid):
        br = ar[pos]
        c += 1
        pos -= 1
 
    # storing the first element as
    # mid value
    br = ar[0]
    c += 1
 
    # if n is odd then store
    # the last value in br[] the
    # transverse till 1st index
    if (n % 2 != 0):
        br = ar[n - 1]
        c += 1
 
    # storing the first element of
    # array as mid value
    while(pos >= 1):
        br = ar[pos]
        c += 1
        pos -= 1
 
    # returning br[] array
    return br
 
# Driver Code
if __name__ == '__main__':
    ar = [2, 8, 5, 9, 10]
    n = len(ar)
 
    # Function Call
    res = rearrange(ar, n)
 
    # Print answer
    for i in range(n):
        print(res[i],end = " ")
         
        # This code is contributed by ipg2016107.

C#




// C# Program of the above approach
using System;
 
public class GFG
{
 
// Function to to shift the middle
// element to the start and end of
// the array alternatively, till
// the middle element becomes equal to
// the first element of the original Array
static int[] rearrange(int[] ar, int n)
{
   
    // creating the array to store
    // rearranged value
    int[] br = new int[n];
 
    // initialising pos to last index
    int pos = n - 1;
 
    // if n is odd then we will
    // transverse the array
    // from second last element
    if (n % 2 != 0)
        pos = pos - 1;
 
    // storing index of middle element
    int mid = n / 2;
 
    // index variable for rearranged array
    int c = 0;
 
    // transversing the array from
    // the pos to mid index
    // and storing it in br[] array
    for (; pos >= mid; pos--)
        br = ar[pos];
 
    // storing the first element as
    // mid value
    br = ar[0];
 
    // if n is odd then store
    // the last value in br[] the
    // transverse till 1st index
    if (n % 2 != 0)
        br = ar[n - 1];
 
    // storing the first element of
    // array as mid value
    for (; pos >= 1; pos--)
        br = ar[pos];
 
    // returning br[] array
    return br;
}
   
// Driver Code
public static void Main(String[] args)
{
    int []ar = { 2, 8, 5, 9, 10 };
    int n = ar.Length;
 
    // Function Call
    int[] res = rearrange(ar, n);
 
    // Print answer
    for (int i = 0; i < n; i++)
        Console.Write(res[i]+ " ");
}
}
 
// This code is contributed by Amit Katiyar

Javascript




<script>
// javascript Program of the above approach
 
// Function to to shift the middle
// element to the start and end of
// the array alternatively, till
// the middle element becomes equal to
// the first element of the original Array
function rearrange(ar , n)
{
   
    // creating the array to store
    // rearranged value
    var br = Array.from({length: n}, (_, i) => 0);
 
    // initialising pos to last index
    var pos = n - 1;
 
    // if n is odd then we will
    // transverse the array
    // from second last element
    if (n % 2 != 0)
        pos = pos - 1;
 
    // storing index of middle element
    var mid = parseInt(n / 2);
 
    // index variable for rearranged array
    var c = 0;
 
    // transversing the array from
    // the pos to mid index
    // and storing it in br array
    for (; pos >= mid; pos--)
        br = ar[pos];
 
    // storing the first element as
    // mid value
    br = ar[0];
 
    // if n is odd then store
    // the last value in br the
    // transverse till 1st index
    if (n % 2 != 0)
        br = ar[n - 1];
 
    // storing the first element of
    // array as mid value
    for (; pos >= 1; pos--)
        br = ar[pos];
 
    // returning br array
    return br;
}
   
// Driver Code
    var ar = [ 2, 8, 5, 9, 10 ];
    var n = ar.length;
 
    // Function Call
    var res = rearrange(ar, n);
 
    // Print answer
    for (var i = 0; i < n; i++)
        document.write(res[i]+ " ");
 
// This code is contributed by Princi Singh
</script>
Output
9 5 2 10 8 

Time Complexity: O(n)
Space Complexity: O(n)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :