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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 the right side array and find the 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 the 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.

Implementation:

C++

 `// C++ program to find ceiling on right side for``// every element.``#include ``using` `namespace` `std;` `void` `closestGreater(``int` `arr[], ``int` `n)``{``    ``set<``int``> s;``    ``vector<``int``> ceilings;` `    ``// Find smallest greater or equal element``    ``// for every array element``    ``for` `(``int` `i = n - 1; i >= 0; i--) {``        ``auto` `greater = s.lower_bound(arr[i]);``        ``if` `(greater == s.end())``            ``ceilings.push_back(-1);``        ``else``            ``ceilings.push_back(*greater);``        ``s.insert(arr[i]);``    ``}` `    ``for` `(``int` `i = n - 1; i >= 0; i--)``        ``cout << ceilings[i] << ``" "``;``}` `int` `main()``{``    ``int` `arr[] = { 50, 20, 200, 100, 30 };``    ``closestGreater(arr, 5);``    ``return` `0;``}`

Java

 `// 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 ts = ``new` `TreeSet();``        ``ArrayList ceilings = ``new` `ArrayList(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);``    ``}``}`

Python3

 `# Python program to find ceiling on right side for``# every element.``import` `bisect` `def` `closestGreater(arr, n):``    ``s ``=` `[]``    ``ceilings ``=` `[]` `    ``# Find smallest greater or equal element``    ``# for every array element``    ``for` `i ``in` `range``(n ``-` `1``, ``-``1``, ``-``1``):``        ``greater ``=` `bisect.bisect_left(s, arr[i])``        ``if` `greater ``=``=` `len``(s):``            ``ceilings.append(``-``1``)``        ``else``:``            ``ceilings.append(s[greater])``        ``s.insert(greater, arr[i])` `    ``for` `i ``in` `range``(n ``-` `1``, ``-``1``, ``-``1``):``        ``print``(ceilings[i], end``=``" "``)` `arr ``=` `[``50``, ``20``, ``200``, ``100``, ``30``]``closestGreater(arr, ``5``)` `# This code is contributed by codebraxnzt`

C#

 `// C# program to find ceiling on right side for``// every element.``using` `System;``using` `System.Collections.Generic;` `public` `class` `TreeSetDemo {``  ``public` `static` `void` `closestGreater(``int``[] arr)``  ``{``    ``int` `n = arr.Length;``    ``SortedSet<``int``> ts = ``new` `SortedSet<``int``>();``    ``List<``int``> ceilings = ``new` `List<``int``>(n);` `    ``// Find smallest greater or equal element``    ``// for every array element``    ``for` `(``int` `i = n - 1; i >= 0; i--) {``      ``int` `greater = lower_bound(ts, arr[i]);``      ``if` `(greater == -1)``        ``ceilings.Add(-1);``      ``else``        ``ceilings.Add(greater);``      ``ts.Add(arr[i]);``    ``}``    ``ceilings.Sort((a,b)=>a-b);``    ``for` `(``int` `i = n - 1; i >= 0; i--)``      ``Console.Write(ceilings[i] + ``" "``);``  ``}``  ``public` `static` `int` `lower_bound(SortedSet<``int``> s, ``int` `val)``  ``{``    ``List<``int``> temp = ``new` `List<``int``>();``    ``temp.AddRange(s);``    ``temp.Sort();``    ``temp.Reverse();` `    ``if` `(temp.IndexOf(val) + 1 == temp.Count)``      ``return` `-1;``    ``else` `if``(temp[temp.IndexOf(val) +1]>val)``      ``return` `-1;``    ``else``      ``return` `temp[temp.IndexOf(val) +1];``  ``}` `  ``public` `static` `void` `Main(String[] args)``  ``{``    ``int``[] arr = { 50, 20, 200, 100, 30 };``    ``closestGreater(arr);``  ``}``}` `// This code is contributed by Rajput-Ji`

Javascript

 `// JavaScript code for the above approach` `function` `closestGreater(arr, n) {``  ``let s = [];``  ``let ceilings = [];` `  ``// Find smallest greater or equal element``  ``// for every array element``  ``for` `(let i = n - 1; i >= 0; i--) {``    ``let greater = bisect_left(s, arr[i]);``    ``if` `(greater == s.length) {``      ``ceilings.push(-1);``    ``} ``else` `{``      ``ceilings.push(s[greater]);``    ``}``    ``s.splice(greater, 0, arr[i]);``  ``}``  ``let temp = [];``  ``for` `(let i = n - 1; i >= 0; i--) {``    ``temp.push(ceilings[i]);``  ``}``  ``console.log(temp.join(``" "``));``}` `function` `bisect_left(s, x) {``  ``let lo = 0;``  ``let hi = s.length;``  ``while` `(lo < hi) {``    ``let mid = Math.floor((lo + hi) / 2);``    ``if` `(s[mid] < x) {``      ``lo = mid + 1;``    ``} ``else` `{``      ``hi = mid;``    ``}``  ``}``  ``return` `lo;``}` `let arr = [50, 20, 200, 100, 30];``closestGreater(arr, 5);`  `// This code is contributed by Prince Kumar`

Output

`100 30 -1 -1 -1 `

Time Complexity: O(n Log n)
Auxiliary Space: O(n)

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