Rearrange 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.

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

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

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

C++

 // C++ program to rearrange an array in minimum // maximum form #include using namespace std;    // Prints max at first position, min at second position // second max at third position, second min at fourth // position and so on. 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 program to test above function int main() {     int arr[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9 };     int n = sizeof(arr) / sizeof(arr);        cout << "Original Arrayn";     for (int i = 0; i < n; i++)         cout << arr[i] << " ";        rearrange(arr, n);        cout << "\nModified Array\n";     for (int i = 0; i < n; i++)         cout << arr[i] << " ";     return 0; }

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("\nModified Array\n");         for (int i = 0; i < n; i++)             System.out.print(arr[i] + " ");     } }    // This code is contributed by Swetank Modi

Python3

 # Python3 program to rearrange an  # array in minimum maximum form    # Prints max at first position, min at second position # second max at third position, second min at fourth # position and so on. def rearrange(arr, n):        # Initialize index of first minimum      # and first maximum element     max_idx = n - 1     min_idx = 0        # Store maximum element of array     max_elem = arr[n-1] + 1        # Traverse array elements     for i in range(0, n) :            # At even index : we have to put maximum element         if i % 2 == 0 :             arr[i] += (arr[max_idx] % max_elem ) * max_elem             max_idx -= 1            # At odd index : we have to put minimum element         else :             arr[i] += (arr[min_idx] % max_elem ) * max_elem             min_idx += 1        # array elements back to it's original form     for i in range(0, n) :         arr[i] = arr[i] / max_elem        # Driver Code arr = [1, 2, 3, 4, 5, 6, 7, 8, 9] n = len(arr)    print ("Original Array")    for i in range(0, n):     print (arr[i], end = " ")        rearrange(arr, n)    print ("\nModified Array") for i in range(0, n):     print (int(arr[i]), end = " ")        # This code is contributed by Shreyanshi Arun.

C#

 // C# program to rearrange an // array in minimum maximum form using System;    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()     {         int[] arr = { 1, 2, 3, 4, 5, 6, 7, 8, 9 };         int n = arr.Length;         Console.WriteLine("Original Array");         for (int i = 0; i < n; i++)             Console.Write(arr[i] + " ");         Console.WriteLine();            rearrange(arr, n);            Console.WriteLine("Modified Array");         for (int i = 0; i < n; i++)             Console.Write(arr[i] + " ");     } }    // This code is contributed by vt_m.

PHP



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:

C++

 // C++ program to rearrange an array in minimum // maximum form #include using namespace std;    // Prints max at first position, min at second position // second max at third position, second min at fourth // position and so on. 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;     // 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 program to test above function int main() {     int arr[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9 };     int n = sizeof(arr) / sizeof(arr);        cout << "Original Array\n";     for (int i = 0; i < n; i++)         cout << arr[i] << " ";        rearrange(arr, n);        cout << "\nModified Array\n";     for (int i = 0; i < n; i++)         cout << arr[i] << " ";     return 0; }

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;         // 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("\nModified Array\n");         for (int i = 0; i < n; i++)             System.out.print(arr[i] + " ");     } }

Python3

 # Python 3 program to rearrange an  # array in minimum maximum form     # Prints max at first position, min  # at second position second max at # third position, second min at  # fourth position and so on.  def rearrange(arr, n):        # initialize index of first minimum      # and first maximum element      max_ele = arr[n - 1]     min_ele = arr        # traverse array elements      for i in range(n):                    # 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 arr = [1, 2, 3, 4, 5, 6, 7, 8, 9] n = len(arr) print("Origianl Array") for i in range(n):     print(arr[i], end = " ")    rearrange(arr, n) print("\nModified Array") for i in range(n):     print(arr[i], end = " ")    # This code is contributed by Shrikant13

C#

 // C# program to rearrange  // an array in minimum  // maximum form using System;    class GFG {     // 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;                    // 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     static public void Main ()     {         int []arr = {1, 2, 3, 4,                       5, 6, 7, 8, 9};         int n = arr.Length;            Console.WriteLine("Original Array");         for (int i = 0; i < n; i++)             Console.Write(arr[i] + " ");            rearrange(arr, n);            Console.Write("\nModified Array\n");         for (int i = 0; i < n; i++)             Console.Write(arr[i] + " ");     } }    // This code is contributed by ajit

Output :

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

Thanks Apollo Doley for suggesting this approach.

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