Smallest Greater Element on Right Side

3Given an array of distinct elements, print the closest greater element for every element. The closest greater element for an element x is the smallest element on the right side of x in array which is greater than x. Elements for which no greater element exist, consider next greater element as -1. 

Examples: 

Input: arr[] = {4, 5, 2, 25}
Output: 
Element NGE
 4 --> 5
 5 --> 25
 2 --> 25
 25 --> -1

Input: arr[] = {4, 10, 7}
Output: 
Element NGE
 4 --> 7
 10 --> -1
 7 --> -1

Approach: In this post, we will be discussing how to find the Next Greater Element using C++ STL(set). 

Finding the smallest greater element on the right side will be like finding the first greater element of the current element in a list that is sorted. 
Consider example 1, The sorted list would look like 2, 4, 5, 25. 

Here for element 4, the greater element is 5 as it is next to it, so we print 5 and remove 4 because it would not be greater to the other elements since it is no longer on anyone's right. 
Similarly, for 5 it is 25 and we remove 5 from the list, as 5 will not be on the right side of 2 or 25, so it can be deleted.

Given below are the steps to find the Next Greater Element of every index element. 

• Insert all the elements in a Set, it will store all the elements in an increasing order.
• Iterate on the array of elements, and for each index, find the upper_bound of the current index element. The upper_bound() returns an iterator which can point to the following position. 
  
  1. If the iterator is pointing to a position past the last element, then there exists no NGE to the current index element.
  2. If the iterator points to a position referring to an element, then that element is the NGE to the current index element.
• Find the position of current index element at every traversal and remove it from the set using >lower_bound() and erase() functions of set.

Implementation:

Element NGE
 4 ----> 5
 5 ----> 25
 2 ----> 25
 25 ----> -1

Complexity Analysis:

• Time Complexity: O(N*logN), as we are using a loop to traverse N times and in each traversal we are inserting to the set which will cost us logN time.
• Auxiliary Space: O(N), as we are using extra space for set ms.