# Next greater element in same order as input

• Difficulty Level : Medium
• Last Updated : 02 Sep, 2022

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.

1. 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.
2. 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
3. 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++

 `// A Stack based C++ program to find next``// greater element for all array elements``// in same order as input.``#include ``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 n-1 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 s = ``new` `Stack();` `    ``int` `arr1[] = ``new` `int``[n];` `    ``// iterating from n-1 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 n-1 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`

## C#

 `// A Stack based C# program to find next``// greater element for all array elements``// in same order as input.``using` `System;``using` `System.Collections.Generic;` `class` `GFG``{` `/* prints element and NGE pair for all``elements of arr[] of size n */``static` `void` `printNGE(``int` `[]arr, ``int` `n)``{``    ``Stack<``int``> s = ``new` `Stack<``int``>();` `    ``int` `[]arr1 = ``new` `int``[n];` `    ``// iterating from n-1 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.Count != 0 && 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.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]);``}` `// Driver Code``public` `static` `void` `Main(String[] args)``{``    ``int` `[]arr = { 11, 13, 21, 3 };``    ``int` `n = arr.Length;``    ``printNGE(arr, n);``}``}` `// This code is contributed by Ajay Kumar`

## Javascript

 ``

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)

My Personal Notes arrow_drop_up