Smallest Greater Element on Right Side (original) (raw)

Last Updated : 11 Jul, 2025

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.

Implementation:

C++ `

// C++ program to print the // NGE's of array elements using // C++ STL #include <bits/stdc++.h> using namespace std;

// Function to print the NGE void printNGE(int a[], int n) { 

set<int> ms; 

// insert in the multiset container 
for (int i = 0; i < n; i++) 
    ms.insert(a[i]); 

cout << "Element "
    << "NGE"; 

// traverse for all array elements 
for (int i = 0; i < n; i++) { 

    // find the upper_bound in set 
    auto it = ms.upper_bound(a[i]); 

    // if points to the end, then 
    // no NGE of that element 
    if (it == ms.end()) { 
        cout << "\n " << a[i] 
            << " ----> " << -1; 
    } 

    // print the element at that position 
    else { 
        cout << "\n " << a[i] 
            << " ----> " << *it; 
    } 

    // find the first occurrence of 
    // the index element and delete it 
    it = ms.lower_bound(a[i]); 

    // delete one occurrence 
    // from the container 
    ms.erase(it); 
}

}

// Driver Code int main() { int a[] = { 4, 5, 2, 25 }; int n = sizeof(a) / sizeof(a[0]); 

// Function call to print the NGE 
printNGE(a, n); 
return 0;

}

Java

// C++ program to print the // NGE's of array elements using import java.util.TreeSet;

class Geeks {

// Function to print the NGE
static void printNGE(int[] a, int n)
{

    // Tree Set is an ordered set used to
    // store elements in a sorted manner
    TreeSet<Integer> t = new TreeSet<>();

    // Adding elements into the set
    for (int i = 0; i < n; i++)
        t.add(a[i]);

    System.out.println("ELEMENT     NGE");

    for (int i = 0; i < n; i++) {

        // If the elements does not have an upper bound
        // or an element greater than it,
        // higher method of TreeSet class will return NULL
        if (t.higher(a[i]) == null)
            System.out.println(a[i] + " ----> "
                               + "-1");

        // Otherwise print the upper bound of that element
        else
            System.out.println(a[i] + " ----> " + t.higher(a[i]));

        // Remove the current element from the set
        t.remove(a[i]);
    }
}

// Driver code
public static void main(String[] args)
{

    int a[] = { 4, 5, 2, 25 };
    int n = a.length;

    printNGE(a, n);
}

}

Python3

Python3 program to print the

NGE's of array elements

from bisect import bisect_right as upper_bound,
bisect_left as lower_bound

Function to print the NGE

def printNGE(a: list, n): ms = set()

# insert in the multiset container
for i in range(n):
    ms.add(a[i])

print("Element NGE", end = "")

# Required because Python sets
# are not sorted
new_arr = list(ms)
new_arr.sort()

# traverse for all array elements
for i in range(n):

    # find the upper_bound in set
    it = upper_bound(new_arr, a[i])

    # if points to the end, then
    # no NGE of that element
    if (it == len(new_arr)):
        print("\n %d ----> -1" % a[i], end = "")

    # print the element at that position
    else:
        print("\n %d ----> %d" % (a[i], 
                new_arr[it]), end = "")

    # find the first occurrence of
    # the index element and delete it
    it = lower_bound(new_arr, a[i])

    # delete one occurrence
    # from the container
    new_arr.remove(new_arr[it])

Driver Code

if name == "main": a = [4, 5, 2, 25] n = len(a)

# Function call to print the NGE
printNGE(a, n)

This code is contributed by

sanjeev2552

C#

// C# program for the above approach

using System; using System.Collections.Generic;

class Geeks { // Function to print the NGE static void printNGE(int[] a, int n) { // insert in the multiset container of array a SortedSet s = new SortedSet(a);

    Console.WriteLine("Element NGE");
        
    // traverse for all array elements 
    for (int i = 0; i < n; i++) {
        SortedSet<int>.Enumerator enumr = s.GetViewBetween(a[i] + 1, int.MaxValue).GetEnumerator();
            
        
        // if points to the end, then 
        // no NGE of that element 
        if (!enumr.MoveNext()) {
            Console.WriteLine($"{a[i]} ----> -1");
        }
        // print the element at that position 
        else {
            Console.WriteLine($"{a[i]} ----> {enumr.Current}");
        }
        
        // delete one occurrence 
        // from the container 
        s.Remove(a[i]);
    }
}
    
    
// Driver Code
public static void Main()
{
    
    int[] a = { 4, 5, 2, 25 };
    int n = a.Length;
    
    // Function call to print the NGE 
    printNGE(a, n);
}

}

// This code is contributed by codebraxnzt

JavaScript

`

Output

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

Complexity Analysis: