Ceiling in right side for every element in an array

• Difficulty Level : Easy
• Last Updated : 23 Jun, 2020

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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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.

C++

 // C++ program to find ceiling on right side for// every element.#include using namespace std;  void closestGreater(int arr[], int n){    set s;    vector 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);    }}
Output:
100 30 -1 -1 -1

Time Complexity : O(n Log n)

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