Inverse of an Array

package jan20;

public class InverseArray {
    public static void main(String[] args) {
        // Create and initialize the input array
        // Time Complexity: O(1)
        // Space Complexity: O(n) – storing input array of size n
        int[] arr = new int[]{4, 0, 2, 3, 1};

        // Create an output array to store the inverse
        // Time Complexity: O(1)
        // Space Complexity: O(n) – storing the result of inversion
        int[] ans = new int[arr.length];

        // Loop through the input array
        // Time Complexity: O(n) – loop runs n times
        // Space Complexity: O(1) – constant space used in loop
        for (int i = 0; i < arr.length; i++) {
            int index = arr[i];     // O(1)
            ans[index] = i;         // O(1)
        }

        // Print the inverse array
        // Time Complexity: O(n) – printing all n elements
        // Space Complexity: O(1)
        for (int x : ans)
            System.out.print(x + "\t");
    }
}

Case
Time Complexity
Explanation
Best Case
O(n)
Every element is processed once.
Average Case
O(n)
All elements are traversed and reassigned.
Worst Case
O(n)
Even if the array is in reverse or random order, all n elements are still processed once.