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 |
Java
// Java program for above approachimport java.util.*;class GFG { // Function for shifting middle element. static void alternateShift(ArrayList<Integer> 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.get(mid) != x) { // pop middle element int z = arr.get(mid); arr.remove(mid); // if c is even then insert z // at start of the array if (c % 2 == 0) { arr.add(0, z); } // if c is odd then insert z // at end of the array else { arr.add(z); } // increment count c c++; } } public static void main(String[] args) { ArrayList<Integer> arr = new ArrayList<>(Arrays.asList(2, 8, 5, 9, 10)); // initialize a to zero index array value int a = arr.get(0); // call AlternateShift function alternateShift(arr, a); // print the changed array Unpacking array for (int i = 0; i < arr.size(); ++i) { System.out.print(arr.get(i) + " "); } }} |
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 += 1Arr = [2, 8, 5, 9, 10]# initialize a to zero index array valuea = Arr[0]# call AlternateShift functionAlternateShift(Arr, a)# print the changed array Unpacking arrayprint(*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> |
C#
// C# program for above approachusing System;using System.Collections.Generic;using System.Linq;public class GFG { // Function for shifting middle element. static void AlternateShift(List<int> arr, int x) { // get middle index int mid = arr.Count / 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.RemoveAt(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.Add(z); } // increment count c c++; } } static public void Main() { // Code List<int> arr = new List<int>{ 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.Count; ++i) { Console.Write(arr[i] + " "); } }}// This code is contributed by lokesh. |
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 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 Arrayint* 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 Codeint 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 approachimport java.util.*;class GFG{// Function 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 Arraystatic 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 Codepublic 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 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 Arraydef 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 Codeif __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 approachusing System;public class GFG{// Function 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 Arraystatic 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 Codepublic 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 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 Arrayfunction 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)
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