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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);`   `    ``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

 `

## 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;` `    ``// 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[``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;` `        `  `        ``// 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);`   `    ``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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