# 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.

`// Java program to find ceiling on right side for ` `// every element. ` `import` `java.util.*; ` ` ` `class` `TreeSetDemo { ` ` ` `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) + ` `" "` `); ` ` ` `} ` ` ` ` ` `public` `static` `void` `main(String[] args) ` ` ` `{ ` ` ` `int` `[] arr = {` `50` `, ` `20` `, ` `200` `, ` `100` `, ` `30` `}; ` ` ` `closestGreater(arr); ` ` ` `} ` `} ` |

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**Output:**

100 30 -1 -1 -1

Time Complexity : O(n Log n)

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