Sort Array such that smallest is at 0th index and next smallest it at last index and so on

Given an array, arr[] of N integers, the task is to rearrange the array elements such that the smallest element is at the 0th position, second smallest element is at the (N-1)th position, the third smallest element at 1st position, 4th smallest element is at the (N-2)th position, and so on for all integers in arr[].

Examples:

Input: arr[] = {10, 23, 12, 17, 9}
Output: 9 12 23 17 10
Explanation:
The smallest element is 9 which is put in the index 0.
Then the second smallest element is 10 which is put in the last position.
The third smallest element is put in the second position from the start.
The fourth smallest in the second position from the last and so on.

Input: arr[] = {1, 3, 3, 4, 5}
Output: 1 3 5 4 3

Approach: We will be using the Two Pointer Technique. The idea is to iterate from the start marked by variable i to end marked by variable j of the array alternatively until they meet in the middle and keep updating the minimum values at these indices. Below are the steps:

Set the element at index i as the minimum value. Iterate over [i, j] and compare each value in this range with arr[i], and if the value in the range is less than arr[i] then swap the value between arr[i] and current element. Increment the value of i.
Set the element at index j as the minimum value. Iterate over [i, j] and compare each value in this range with arr[j], and if the value in the range is less than arr[j] then swap the value between arr[j] and current element. Decrement the value of j.
Place the smallest element in the i^th position at the first iteration and the next smallest element in the j^th position. Do this until both i and j points to the same position.

Below is the implementation of the above approach:

Java

// Java program for the above approach 
import java.io.*; 
import java.util.*; 
  
public class GFG { 
  
    // Function to perform the rearrangement 
    static void rearrange(int[] a) 
    { 
        // Initialize variables 
        int i = 0, j = a.length - 1; 
        int min = 0, k, x = 0, temp; 
  
        // Loop until i crosses j 
        while (i < j) { 
  
            // This check is to find the 
            // minimum values in the 
            // ascending order 
  
            for (k = i; k <= j; k++) { 
                if (a[k] < a[min]) 
                    min = k; 
            } 
  
            // Condition to alternatively 
            // iterate variable i and j 
            if (x % 2 == 0) { 
  
                // Perform swap operation 
                temp = a[i]; 
                a[i] = a[min]; 
                a[min] = temp; 
  
                // Increment i 
                i++; 
  
                // Assign the value of min 
                min = i; 
            } 
            else { 
  
                // Perform swap 
                temp = a[j]; 
                a[j] = a[min]; 
                a[min] = temp; 
  
                // Decrement i 
                j--; 
  
                // Assign the value of min 
                min = j; 
            } 
            x++; 
        } 
  
        // Print the array 
        for (i = 0; i < a.length; i++) 
            System.out.print(a[i] + " "); 
    } 
  
    // Driver Code 
    public static void main(String[] args) 
    { 
        // Given Array arr[] 
        int arr[] = { 1, 3, 3, 4, 5 }; 
  
        // Function Call 
        rearrange(arr); 
    } 
}

Output:

1 3 5 4 3

Time Complexity: O(N²)
Auxiliary Space: O(1)

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Java

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