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.

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 <bits/stdc++.h>
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.

PHP

<?php
// PHP 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.
function 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 = 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 ($i = 0; $i < $n; $i++)
        $arr[$i] = (int)($arr[$i] / $max_elem);
}

// Driver Code
$arr = array(1, 2, 3, 4, 5, 6, 7, 8, 9);
$n = sizeof($arr);

echo "Original Array" . "\n";
for ($i = 0; $i < $n; $i++)
    echo $arr[$i] . " ";

rearrange($arr, $n);

echo "\nModified Array\n";
for ($i = 0; $i < $n; $i++)
    echo $arr[$i] . " ";

// This code is contributed
// by Akanksha Rai(Abby_akku)

Javascript

<script>

// JavaScript 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.
function rearrange(arr, n)
{
    // initialize index of first minimum and first
    // maximum element
    let max_idx = n - 1, min_idx = 0;

    // store maximum element of array
    let max_elem = arr[n - 1] + 1;

    // traverse array elements
    for (let 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 (let i = 0; i < n; i++)
        arr[i] = Math.floor(arr[i] / max_elem);
}

// Driver program to test above function

    let arr = [ 1, 2, 3, 4, 5, 6, 7, 8, 9 ];
    let n = arr.length;

    document.write("Original Array<br>");
    for (let i = 0; i < n; i++)
        document.write(arr[i] + " ");

    rearrange(arr, n);

    document.write("<br>Modified Array<br>");
    for (let i = 0; i < n; i++)
        document.write(arr[i] + " ");

// This code is contributed by Surbhi Tyagi.

</script>

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 <bits/stdc++.h>
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

<script>

// Javascript 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.
    function rearrange(arr, n)
    {
        // initialize index of first minimum and first
        // maximum element
        let max_ele = arr[n - 1];
        let min_ele = arr[0];
        // traverse array elements
        for (let 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

    let arr = [ 1, 2, 3, 4, 5, 6, 7, 8, 9 ];
    let n = arr.length;

    document.write("Original Array" + "<br />");
    for (let i = 0; i < n; i++)
        document.write(arr[i] + " ");
    document.write("<br />");

    rearrange(arr, n);

    document.write("\nModified Array\n" + "<br />");
    for (let i = 0; i < n; i++)
       document.write(arr[i] + " ");

</script>

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 <iostream>
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. 

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