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

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 elementstep1: 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 :Python3def AlternateShift(arr, x):          mid = len(arr) // 2          c = 0              while arr[mid] != x:                  z = arr.pop(mid)                          if c % 2 == 0:            arr.insert(0, z)                          else:            arr.append(z)                  c += 1    Arr = [2, 8, 5, 9, 10]  a = Arr  AlternateShift(Arr, a)  print(*Arr)Output9 5 2 10 8Time 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++#include using namespace std;  int* rearrange(int* ar, int n){            int* br = new int[n];          int pos = n – 1;                  if (n % 2 != 0)        pos = pos – 1;          int mid = n / 2;          int c = 0;                  for (; pos >= mid; pos–)        br[c++] = ar[pos];              br[c++] = ar;                  if (n % 2 != 0)        br[c++] = ar[n – 1];              for (; pos >= 1; pos–)        br[c++] = ar[pos];          return br;}int main(){    int ar[] = { 2, 8, 5, 9, 10 };    int n = sizeof(ar) / sizeof(ar);          int* res = rearrange(ar, n);          for (int i = 0; i < n; i++)        cout 