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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 Practice

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 the 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[0]);       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.



Javascript



Output

Original Arrayn1 2 3 4 5 6 7 8 9
Modified Array
9 1 8 2 7 3 6 4 5

Time Complexity: O(n)
Auxiliary Space: O(1), as no extra space is used

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[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 program to test above function int main() {     int arr[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9 };     int n = sizeof(arr) / sizeof(arr[0]);       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[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("\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[0]       # 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("Original 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[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     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

Javascript



Output

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

Another Approach:

C++

 #include using namespace std;   int main() {     int a[] = { 11, 12, 13, 14, 15, 16 };     int n = sizeof(a) / sizeof(a[0]);       int last[n];     int min = 0, max = n - 1;     int count = 0;     for (int i = 0; min <= max; i++) {         if (count % 2 == 0) {             last[i] = a[max];             max--;         }         else {             last[i] = a[min];             min++;         }         count++;     }     for (int i = 0; i < n; i++)         cout << last[i] << " ";       return 0; }

Java

 import java.util.Arrays;   // Defining the class public class Main {     // Main function     public static void main(String[] args)     {         // Initializing an array a         int[] a = { 11, 12, 13, 14, 15, 16 };         // Finding the length of array a         int n = a.length;         // Initializing an array last with size n         int[] last = new int[n];         // Initializing variables min and max           int min_val = 0;         int max_val = n - 1;         // Initializing a variable count to keep track of         // iterations           int count = 0;         for (int i = 0; i < n; i++) {             // If count is even, store the value of             // a[max_val] in last[i] and decrement max_val             if (count % 2 == 0) {                 last[i] = a[max_val];                 max_val -= 1;             }             // If count is odd, store the value of             // a[min_val] in last[i] and increment min_val             else {                 last[i] = a[min_val];                 min_val += 1;             }             // Increment the value of count             count += 1;         }           // Printing the values in array last         System.out.println(Arrays.toString(last));     } }

Python3

 # Initializing an array a a = [11, 12, 13, 14, 15, 16]   # Finding the length of array a n = len(a)   # Initializing an array last with size n last = [0] * n   # Initializing variables min and max min_val = 0 max_val = n - 1   # Initializing a variable count to keep track of iterations count = 0   # Looping through the array for i in range(n):     # If count is even, store the value of     #a[max_val] in last[i] and decrement max_val     if count % 2 == 0:         last[i] = a[max_val]         max_val -= 1     # If count is odd, store the value of     # a[min_val] in last[i] and increment min_val     else:         last[i] = a[min_val]         min_val += 1       # Increment the value of count     count += 1   # Printing the values in array last for i in range(n):     print(last[i], end=' ')

C#

 // C# program to rearrange // an array in minimum // maximum form using System;   class GFG {     static public void Main ()     {         int[] a = { 11, 12, 13, 14, 15, 16 };         int n = a.Length;               int[] last = new int[n];         int min = 0, max = n - 1;         int count = 0;         for (int i = 0; min <= max; i++) {             if (count % 2 == 0) {                 last[i] = a[max];                 max--;             }             else {                 last[i] = a[min];                 min++;             }             count++;         }         for (int i = 0; i < n; i++)             Console.Write(last[i] + " ");           } }

Javascript

 // Initializing an array a let a = [11, 12, 13, 14, 15, 16];   // Finding the length of array a let n = a.length;   // Initializing an array last with size n let last = new Array(n).fill(0);   // Initializing variables min and max let min_val = 0; let max_val = n - 1;   // Initializing a variable count to keep track of iterations let count = 0;   // Looping through the array for (let i = 0; i < n; i++) { // If count is even, store the value of //a[max_val] in last[i] and decrement max_val if (count % 2 == 0) { last[i] = a[max_val]; max_val--; } // If count is odd, store the value of // a[min_val] in last[i] and increment min_val else { last[i] = a[min_val]; min_val++; }   // Increment the value of count count++;   }   // Printing the values in array last console.log(last.join(' '));   // This code is contributed by adityash4x71

Output

16 11 15 12 14 13

Time Complexity: O(n), where n is the size of the array
Auxiliary Space: O(1), as no extra space is used

Thanks Apollo Doley for suggesting this approach.

This article is contributed by Aarti_Rathi Nishant Singh . If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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