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 1Input : 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 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 from the current element and that element will be the answer for current element
 If stack gets empty while doing the pop operation then the answer would be 1
Then we will stored the answer in an array for the current index.
Below is the implementation of above approach:
C++
// A Stack based C++ program to find next // greater element for all array elements // in same order as input. #include <bits/stdc++.h> using namespace std; /* prints element and NGE pair for all elements of arr[] of size n */ void printNGE( int arr[], int n) { stack< int > s; int arr1[n]; // iterating from n1 to 0 for ( int i = n  1; i >= 0; i) { /*We will pop till we get the greater element on top or stack gets empty*/ while (!s.empty() && s.top() < arr[i]) s.pop(); /*if stack gots empty means there is no element on right which is greater than the current element. if not empty then the next greater element is on top of stack*/ 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; } /* Driver program to test above functions */ int main() { int arr[] = { 11, 13, 21, 3 }; int n = sizeof (arr) / sizeof (arr[0]); printNGE(arr, n); return 0; } 
Java
// A Stack based Java program to find next // greater element for all array elements // in same order as input. import java.util.*; class GfG { /* prints element and NGE pair for all elements of arr[] of size n */ static void printNGE( int arr[], int n) { Stack<Integer> s = new Stack<Integer>(); int arr1[] = new int [n]; // iterating from n1 to 0 for ( int i = n  1 ; i >= 0 ; i) { /*We will pop till we get the greater element on top or stack gets empty*/ while (!s.isEmpty() && s.peek() < arr[i]) s.pop(); /*if stack gots empty means there is no element on right which is greater than the current element. if not empty then the next greater element is on top of stack*/ 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]); } /* Driver program to test above functions */ public static void main(String[] args) { int arr[] = { 11 , 13 , 21 , 3 }; int n = arr.length; printNGE(arr, n); } } 
Python3
# A Stack based Python3 program to find next
# greater element for all array elements
# in same order as input.
# prints element and NGE pair for all
# elements of arr[] of size n
def printNGE(arr, n):
s = list()
arr1 = [0 for i in range(n)]
# iterating from n1 to 0
for i in range(n – 1, 1, 1):
# We will pop till we get the greater
# element on top or stack gets empty
while (len(s) > 0 and s[1] < arr[i]):
s.pop()
# if stack gots empty means there
# is no element on right which is
# greater than the current element.
# if not empty then the next greater
# element is on top of stack
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] )
# Driver Code
arr = [ 11, 13, 21, 3 ]
n = len(arr)
printNGE(arr, n)
# This code is contributed by Mohit kumar 29
Output :
11  13 13  21 21  1 3  1
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)
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