Ceiling in right side for every element in an array

http://massivealgorithms.blogspot.com/2015/12/leetcode-315-count-of-smaller-numbers.html
https://www.geeksforgeeks.org/ceiling-in-right-side-for-every-element-in-an-array/

Given an array of integers, find the closest greater element for every element. If there is no greater element then print -1

Examples:

Input : arr[] = {10, 5, 11, 10, 20, 12}
Output : 10 10 12 12 -1 -1

Input : arr[] = {50, 20, 200, 100, 30}
Output : 100 30 -1 -1 -1

A simple solution is to run two nested loops. We pick an outer element one by one. For every picked element, we traverse right side array and find closest greater or equal element. Time complexity of this solution is O(n*n)

A better solution is to use sorting. We sort all elements, then for every element, traverse toward right until we find a greater element (Note that there can be multiple occurrences of an element).

An efficient solution is to use Self Balancing BST (Implemented as set in C++ and TreeSet in Java). In a Self Balancing BST, we can do both insert and ceiling operations in O(Log n) time.

    public static void closestGreater(int[] arr)
    {
        int n = arr.length;
        TreeSet<Integer> ts = new TreeSet<Integer>();
        ArrayList<Integer> ceilings = new ArrayList<Integer>(n);
 
        // Find smallest greater or equal element
        // for every array element
        for (int i = n - 1; i >= 0; i--) {
            Integer greater = ts.ceiling(arr[i]);
            if (greater == null)
                ceilings.add(-1);
            else
                ceilings.add(greater);
            ts.add(arr[i]);
        }
 
        for (int i=n-1; i>=0; i--)
           System.out.print(ceilings.get(i) + " ");
    }