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

06 Jan, 2022
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 formpublic 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 
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 formpublic 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 
Please refer complete article on Rearrange an array in maximum minimum form | Set 2 (O(1) extra space) for more details!
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