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# Rearrange an array in maximum minimum form using Two Pointer Technique

Given a sorted array of positive integers, rearrange the array alternately i.e first element should be a maximum value, at second position minimum value, at third position second max, at fourth position 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}

## Rearrange an array in maximum minimum form using Two Pointers:

The idea is to use an auxiliary array. We maintain two pointers one to the leftmost or smallest element and the other to the rightmost or largest element. We move both pointers toward each other and alternatively copy elements at these pointers to an auxiliary array. Finally, we copy the auxiliary array back to the original array.

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){    // Auxiliary array to hold modified array    int temp[n];      // Indexes of smallest and largest elements    // from remaining array.    int small = 0, large = n - 1;      // To indicate whether we need to copy remaining    // largest or remaining smallest at next position    int flag = true;      // Store result in temp[]    for (int i = 0; i < n; i++) {        if (flag)            temp[i] = arr[large--];        else            temp[i] = arr[small++];          flag = !flag;    }      // Copy temp[] to arr[]    for (int i = 0; i < n; i++)        arr[i] = temp[i];}  // Driver codeint main(){    int arr[] = { 1, 2, 3, 4, 5, 6 };    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  import java.util.Arrays;  public class GFG {    // Prints max at first position, min at second position    // second max at third position, second min at fourth    // position and so on.    static void rearrange(int[] arr, int n)    {        // Auxiliary array to hold modified array        int temp[] = arr.clone();          // Indexes of smallest and largest elements        // from remaining array.        int small = 0, large = n - 1;          // To indicate whether we need to copy remaining        // largest or remaining smallest at next position        boolean flag = true;          // Store result in temp[]        for (int i = 0; i < n; i++) {            if (flag)                arr[i] = temp[large--];            else                arr[i] = temp[small++];              flag = !flag;        }    }      // Driver code    public static void main(String[] args)    {        int arr[] = new int[] { 1, 2, 3, 4, 5, 6 };          System.out.println("Original Array ");        System.out.println(Arrays.toString(arr));          rearrange(arr, arr.length);          System.out.println("Modified Array ");        System.out.println(Arrays.toString(arr));    }}

## Python3

 # Python 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):    # Auxiliary array to hold modified array    temp = n*[None]      # Indexes of smallest and largest elements    # from remaining array.    small, large = 0, n-1      # To indicate whether we need to copy remaining    # largest or remaining smallest at next position    flag = True      # Store result in temp[]    for i in range(n):        if flag is True:            temp[i] = arr[large]            large -= 1        else:            temp[i] = arr[small]            small += 1          flag = bool(1-flag)      # Copy temp[] to arr[]    for i in range(n):        arr[i] = temp[i]    return arr    # Driver codearr = [1, 2, 3, 4, 5, 6]n = len(arr)print("Original Array")print(arr)print("Modified Array")print(rearrange(arr, n))  # This code is contributed by Pratik Chhajer

## C#

 // C# program to rearrange// an array in minimum// maximum formusing 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.    static void rearrange(int[] arr, int n)    {        // Auxiliary array to        // hold modified array        int[] temp = new int[n];          // Indexes of smallest        // and largest elements        // from remaining array.        int small = 0, large = n - 1;          // To indicate whether we        // need to copy remaining        // largest or remaining        // smallest at next position        bool flag = true;          // Store result in temp[]        for (int i = 0; i < n; i++) {            if (flag)                temp[i] = arr[large--];            else                temp[i] = arr[small++];              flag = !flag;        }          // Copy temp[] to arr[]        for (int i = 0; i < n; i++)            arr[i] = temp[i];    }      // Driver Code    static void Main()    {        int[] arr = { 1, 2, 3, 4, 5, 6 };          Console.WriteLine("Original Array");        for (int i = 0; i < arr.Length; i++)            Console.Write(arr[i] + " ");          rearrange(arr, arr.Length);          Console.WriteLine("\nModified Array");        for (int i = 0; i < arr.Length; i++)            Console.Write(arr[i] + " ");    }}  // This code is contributed// by Sam007



## Javascript



Output

Original Array
1 2 3 4 5 6
Modified Array
6 1 5 2 4 3

Time Complexity: O(N), Iterating over the array of size N 2 times.
Auxiliary Space: O(N), since N extra space has been taken.

Exercise: How to solve this problem if extra space is not allowed?
Rearrange an array in maximum minimum form | Set 2 (O(1) extra space)

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