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.
- If the iterator is pointing to a position past the last element, then there exists no NGE to the current index element.
- 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:
C++
#include <bits/stdc++.h>
using namespace std;
void printNGE( int a[], int n)
{
set< int > ms;
for ( int i = 0; i < n; i++)
ms.insert(a[i]);
cout << "Element "
<< "NGE" ;
for ( int i = 0; i < n; i++) {
auto it = ms.upper_bound(a[i]);
if (it == ms.end()) {
cout << "\n " << a[i]
<< " ----> " << -1;
}
else {
cout << "\n " << a[i]
<< " ----> " << *it;
}
it = ms.lower_bound(a[i]);
ms.erase(it);
}
}
int main()
{
int a[] = { 4, 5, 2, 25 };
int n = sizeof (a) / sizeof (a[0]);
printNGE(a, n);
return 0;
}
|
Java
import java.util.TreeSet;
class Geeks {
static void printNGE( int [] a, int n)
{
TreeSet<Integer> t = new TreeSet<>();
for ( int i = 0 ; i < n; i++)
t.add(a[i]);
System.out.println( "ELEMENT NGE" );
for ( int i = 0 ; i < n; i++) {
if (t.higher(a[i]) == null )
System.out.println(a[i] + " ----> "
+ "-1" );
else
System.out.println(a[i] + " ----> " + t.higher(a[i]));
t.remove(a[i]);
}
}
public static void main(String[] args)
{
int a[] = { 4 , 5 , 2 , 25 };
int n = a.length;
printNGE(a, n);
}
}
|
Python3
from bisect import bisect_right as upper_bound, \
bisect_left as lower_bound
def printNGE(a: list , n):
ms = set ()
for i in range (n):
ms.add(a[i])
print ( "Element NGE" , end = "")
new_arr = list (ms)
new_arr.sort()
for i in range (n):
it = upper_bound(new_arr, a[i])
if (it = = len (new_arr)):
print ( "\n %d ----> -1" % a[i], end = "")
else :
print ( "\n %d ----> %d" % (a[i],
new_arr[it]), end = "")
it = lower_bound(new_arr, a[i])
new_arr.remove(new_arr[it])
if __name__ = = "__main__" :
a = [ 4 , 5 , 2 , 25 ]
n = len (a)
printNGE(a, n)
|
C#
using System;
using System.Collections.Generic;
class Geeks
{
static void printNGE( int [] a, int n)
{
SortedSet< int > s = new SortedSet< int >(a);
Console.WriteLine( "Element NGE" );
for ( int i = 0; i < n; i++) {
SortedSet< int >.Enumerator enumr = s.GetViewBetween(a[i] + 1, int .MaxValue).GetEnumerator();
if (!enumr.MoveNext()) {
Console.WriteLine($ "{a[i]} ----> -1" );
}
else {
Console.WriteLine($ "{a[i]} ----> {enumr.Current}" );
}
s.Remove(a[i]);
}
}
public static void Main()
{
int [] a = { 4, 5, 2, 25 };
int n = a.Length;
printNGE(a, n);
}
}
|
Javascript
<script>
function printNGE(a , n) {
var t = new Set();
for ( var i = 0; i < n; i++)
t.add(a[i]);
document.write( "ELEMENT NGE<br/>" );
for (i = 0; i < n; i++) {
if (upper_bound(t,a[i]) == null )
document.write(a[i] + " ----> " + "-1" + "<br/>" );
else
document.write(a[i] + " ----> " + upper_bound(t,a[i])+ "<br/>" );
t. delete (a[i]);
}
}
function upper_bound(s, val)
{
let temp = [...s];
temp.sort((a, b) => b - a);
return temp[temp.indexOf(val) + 1];
}
var a = [ 4, 5, 2, 25 ];
var n = a.length;
printNGE(a, n);
</script>
|
Output
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.
Last Updated :
14 Mar, 2023
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