Smallest Greater Element on Right Side

• Difficulty Level : Medium
• Last Updated : 11 Dec, 2019

Given 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

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

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.

Below is the implementation of the above approach.

C/C++

 // C++ program to print the// NGE's of array elements using// C++ STL#include using namespace std;  // Function to print the NGEvoid printNGE(int a[], int n){      set 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 Codeint main(){    int a[] = { 4, 5, 2, 25 };    int n = sizeof(a) / sizeof(a);      // Function call to print the NGE    printNGE(a, n);    return 0;}

Java

 // C++ program to print the// NGE's of array elements usingimport 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 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 elementsfrom bisect import bisect_right as upper_bound, \                   bisect_left as lower_bound  # Function to print the NGEdef 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 Codeif __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
Output:
Element   NGE
4 ----> 5
5 ----> 25
2 ----> 25
25 ----> -1

Time Complexity: O(N log N)

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