GeeksforGeeks App
Open App
Browser
Continue

# Java Program For Rearranging An Array In Maximum Minimum Form – Set 2 (O(1) extra space)

Given a sorted array of positive integers, rearrange the array alternately i.e first element should be the maximum value, second minimum value, third-second max, fourth-second min and so on.
Examples:

Input: arr[] = {1, 2, 3, 4, 5, 6, 7}
Output: arr[] = {7, 1, 6, 2, 5, 3, 4}
Input: arr[] = {1, 2, 3, 4, 5, 6}
Output: arr[] = {6, 1, 5, 2, 4, 3}

We have discussed a solution in below post:
Rearrange an array in maximum minimum form | Set 1 : The solution discussed here requires extra space, how to solve this problem with O(1) extra space.

In this post a solution that requires O(n) time and O(1) extra space is discussed. The idea is to use multiplication and modular trick to store two elements at an index.

```even index : remaining maximum element.
odd index  : remaining minimum element.

max_index : Index of remaining maximum element
(Moves from right to left)
min_index : Index of remaining minimum element
(Moves from left to right)

Initialize: max_index = 'n-1'
min_index = 0

// can be any element which is more
// than the maximum value in array
max_element = arr[max_index] + 1

For i = 0 to n-1
If 'i' is even
arr[i] += arr[max_index] % max_element * max_element
max_index--

// if 'i' is odd
ELSE
arr[i] +=  arr[min_index] % max_element * max_element
min_index++```

How does expression “arr[i] += arr[max_index] % max_element * max_element” work ?
The purpose of this expression is to store two elements at index arr[i]. arr[max_index] is stored as multiplier and “arr[i]” is stored as remainder. For example in {1 2 3 4 5 6 7 8 9}, max_element is 10 and we store 91 at index 0. With 91, we can get original element as 91%10 and new element as 91/10.
Below implementation of above idea:

## Java

 `// Java program to rearrange an``// array in minimum maximum form``public` `class` `Main {` `    ``// Prints max at first position, min``    ``// at second position second max at``    ``// third position, second min at``    ``// fourth position and so on.``    ``public` `static` `void` `rearrange(``int` `arr[],``                                 ``int` `n)``    ``{``        ``// Initialize index of first minimum``        ``// and first maximum element``        ``int` `max_idx = n - ``1``, min_idx = ``0``;` `        ``// Store maximum element of array``        ``int` `max_elem = arr[n - ``1``] + ``1``;` `        ``// Traverse array elements``        ``for` `(``int` `i = ``0``; i < n; i++)``        ``{``            ``// At even index : we have to put``            ``// maximum element``            ``if` `(i % ``2` `== ``0``)``            ``{``                ``arr[i] += ((arr[max_idx] % max_elem) *``                            ``max_elem);``                ``max_idx--;``            ``}` `            ``// At odd index : we have to put``            ``// minimum element``            ``else``            ``{``                ``arr[i] += ((arr[min_idx] % max_elem) *``                            ``max_elem);``                ``min_idx++;``            ``}``        ``}` `        ``// Array elements back to it's``        ``// original form``        ``for` `(``int` `i = ``0``; i < n; i++)``            ``arr[i] = arr[i] / max_elem;``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String args[])``    ``{``        ``int` `arr[] = {``1``, ``2``, ``3``, ``4``, ``5``,``                     ``6``, ``7``, ``8``, ``9``};``        ``int` `n = arr.length;` `        ``System.out.println(``        ``"Original Array"``);``        ``for` `(``int` `i = ``0``; i < n; i++)``            ``System.out.print(arr[i] + ``" "``);` `        ``rearrange(arr, n);` `        ``System.out.print(``        ``"Modified Array"``);``        ``for` `(``int` `i = ``0``; i < n; i++)``            ``System.out.print(arr[i] + ``" "``);``    ``}``}``// This code is contributed by Swetank Modi`

Output :

```Original Array
1 2 3 4 5 6 7 8 9
Modified Array
9 1 8 2 7 3 6 4 5 ```

Time Complexity: O(N), as we are using a loop to traverse N times.

Auxiliary Space: O(1), as we are not using any extra space.

Thanks Saurabh Srivastava and Gaurav Ahirwar for suggesting this approach.
Another Approach: A simpler approach will be to observe indexing positioning of maximum elements and minimum elements. The even index stores maximum elements and the odd index stores the minimum elements. With every increasing index, the maximum element decreases by one and the minimum element increases by one. A simple traversal can be done and arr[] can be filled in again.
Note: This approach is only valid when elements of given sorted array are consecutive i.e., vary by one unit.
Below is the implementation of the above approach:

## Java

 `// Java program to rearrange an``// array in minimum maximum form``public` `class` `Main``{``    ``// Prints max at first position, min``    ``// at second position second max at``    ``// third position, second min at``    ``// fourth position and so on.``    ``public` `static` `void` `rearrange(``int` `arr[],``                                 ``int` `n)``    ``{``        ``// Initialize index of first minimum``        ``// and first maximum element``        ``int` `max_ele = arr[n - ``1``];``        ``int` `min_ele = arr[``0``];` `        ``// Traverse array elements``        ``for` `(``int` `i = ``0``; i < n; i++)``        ``{``            ``// At even index : we have to put``            ``// maximum element``            ``if` `(i % ``2` `== ``0``)``            ``{``                ``arr[i] = max_ele;``                ``max_ele -= ``1``;``            ``}` `            ``// At odd index : we have to put``            ``// minimum element``            ``else``            ``{``                ``arr[i] = min_ele;``                ``min_ele += ``1``;``            ``}``        ``}``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String args[])``    ``{``        ``int` `arr[] = {``1``, ``2``, ``3``, ``4``, ``5``,``                     ``6``, ``7``, ``8``, ``9``};``        ``int` `n = arr.length;` `        ``System.out.println(``"Original Array"``);``        ``for` `(``int` `i = ``0``; i < n; i++)``            ``System.out.print(arr[i] + ``" "``);` `        ``rearrange(arr, n);` `        ``System.out.print(``        ``"Modified Array"``);``        ``for` `(``int` `i = ``0``; i < n; i++)``            ``System.out.print(arr[i] + ``" "``);``    ``}``}`

Output :

```Original Array
1 2 3 4 5 6 7 8 9
Modified Array
9 1 8 2 7 3 6 4 5 ```

Time Complexity: O(N), as we are using a loop to traverse N times.

Auxiliary Space: O(1), as we are not using any extra space.

Please refer complete article on Rearrange an array in maximum minimum form | Set 2 (O(1) extra space) for more details!

My Personal Notes arrow_drop_up