Next greater element in same order as input
Given an array, print the Next Greater Element (NGE) for every element. The Next greater Element for an element x is the first greater element on the right side of x in array. Elements for which no greater element exist, consider next greater element as -1. The next greater elements should be printed in same order as input array.
Examples:
Input : arr[] = [4, 5, 2, 25}
Output : 5 25 25 -1
Input : arr[] = [4, 5, 2, 25, 10}
Output : 5 25 25 -1 -1
We have discussed a solution here that does not print same order. Here we traverse array from rightmost element.
- In this approach, we have started iterating from the last element(nth) to the first(1st) element
The benefit is that when we arrive at a certain index his next greater element will be already in the stack, and we can directly get this element while at the same index.
- After reaching a certain index we will pop the stack till we get the greater element on top of the current element and that element will be the answer for current element
- If the stack gets empty while doing the pop operation then the answer would be -1
Then we will store the answer in an array for the current index.
Implementation:
C++
#include <bits/stdc++.h>
using namespace std;
void printNGE( int arr[], int n)
{
stack< int > s;
int arr1[n];
for ( int i = n - 1; i >= 0; i--)
{
while (!s.empty() && s.top() <= arr[i])
s.pop();
if (s.empty())
arr1[i] = -1;
else
arr1[i] = s.top();
s.push(arr[i]);
}
for ( int i = 0; i < n; i++)
cout << arr[i] << " ---> " << arr1[i] << endl;
}
int main()
{
int arr[] = { 11, 13, 21, 3 };
int n = sizeof (arr) / sizeof (arr[0]);
printNGE(arr, n);
return 0;
}
|
Java
import java.util.*;
class GfG {
static void printNGE( int arr[], int n)
{
Stack<Integer> s = new Stack<Integer>();
int arr1[] = new int [n];
for ( int i = n - 1 ; i >= 0 ; i--)
{
while (!s.isEmpty() && s.peek() <= arr[i])
s.pop();
if (s.empty())
arr1[i] = - 1 ;
else
arr1[i] = s.peek();
s.push(arr[i]);
}
for ( int i = 0 ; i < n; i++)
System.out.println(arr[i] + " ---> " + arr1[i]);
}
public static void main(String[] args)
{
int arr[] = { 11 , 13 , 21 , 3 };
int n = arr.length;
printNGE(arr, n);
}
}
|
Python3
def printNGE(arr, n):
s = list ()
arr1 = [ 0 for i in range (n)]
for i in range (n - 1 , - 1 , - 1 ):
while ( len (s) > 0 and s[ - 1 ] < = arr[i]):
s.pop()
if ( len (s) = = 0 ):
arr1[i] = - 1
else :
arr1[i] = s[ - 1 ]
s.append(arr[i])
for i in range (n):
print (arr[i], " ---> " , arr1[i] )
arr = [ 11 , 13 , 21 , 3 ]
n = len (arr)
printNGE(arr, n)
|
C#
using System;
using System.Collections.Generic;
class GFG
{
static void printNGE( int []arr, int n)
{
Stack< int > s = new Stack< int >();
int []arr1 = new int [n];
for ( int i = n - 1; i >= 0; i--)
{
while (s.Count != 0 && s.Peek() <= arr[i])
s.Pop();
if (s.Count == 0)
arr1[i] = -1;
else
arr1[i] = s.Peek();
s.Push(arr[i]);
}
for ( int i = 0; i < n; i++)
Console.WriteLine(arr[i] + " ---> " +
arr1[i]);
}
public static void Main(String[] args)
{
int []arr = { 11, 13, 21, 3 };
int n = arr.Length;
printNGE(arr, n);
}
}
|
Javascript
<script>
function printNGE(arr, n)
{
let s = [];
let arr1 = new Array(n);
for (let i = n - 1; i >= 0; i--)
{
while (!s.length == 0 &&
s[s.length - 1] <= arr[i])
s.pop();
if (s.length == 0)
arr1[i] = -1;
else
arr1[i] = s[s.length - 1];
s.push(arr[i]);
}
for (let i = 0; i < n; i++)
document.write(arr[i] + " ---> " +
arr1[i] + "<br>" );
}
let arr = [ 11, 13, 21, 3 ];
let n = arr.length;
printNGE(arr, n);
</script>
|
Output
11 ---> 13
13 ---> 21
21 ---> -1
3 ---> -1
Complexity Analysis:
- Time Complexity: O(n)
- Auxiliary Space: O(n) There is no extra space required if you want to print the next greater of each element in reverse order of input(means first, for the last element and then for second last and so on till the first element)
Last Updated :
02 Sep, 2022
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