Next Greater Element (NGE) for every element in given Array

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 the array. Elements for which no greater element exist, consider the next greater element as -1. 

Example: 

Input: arr[] = [ 4 , 5 , 2 , 25 ]
Output:  4      –>   5
               5      –>   25
               2      –>   25
              25     –>   -1
Explanation: except 25 every element has an element greater than them present on the right side

Input: arr[] = [ 13 , 7, 6 , 12 ]
Output:  13      –>    -1
                7       –>     12
                6       –>     12
               12      –>     -1
Explanation: 13 and 12 don’t have any element greater than them present on the right side

The idea is to use two loops , The outer loop picks all the elements one by one. The inner loop looks for the first greater element for the element picked by the outer loop. If a greater element is found then that element is printed as next, otherwise, -1 is printed.

Follow the steps mentioned below to implement the idea:

• Traverse the array from index 0 to end.
  • For each element start another loop from index i+1 to end.
    • If a greater element is found in the second loop then print it and break the loop, else print -1.  

Below is the implementation of the above approach:

C++

// Simple C++ program to print
// next greater elements in a
// given array
#include <iostream>
using namespace std;

/* prints element and NGE pair
for all elements of arr[] of size n */
void printNGE(int arr[], int n)
{
    int next, i, j;
    for (i = 0; i < n; i++) {
        next = -1;
        for (j = i + 1; j < n; j++) {
            if (arr[i] < arr[j]) {
                next = arr[j];
                break;
            }
        }
        cout << arr[i] << " --> " << next << endl;
    }
}

// Driver Code
int main()
{
    int arr[] = { 11, 13, 21, 3 };
    int n = sizeof(arr) / sizeof(arr[0]);
    printNGE(arr, n);
    return 0;
}

// This code is contributed
// by Akanksha Rai(Abby_akku)

C

// Simple C program to print next greater elements
// in a given array
#include <stdio.h>

/* prints element and NGE pair for all elements of
arr[] of size n */
void printNGE(int arr[], int n)
{
    int next, i, j;
    for (i = 0; i < n; i++) {
        next = -1;
        for (j = i + 1; j < n; j++) {
            if (arr[i] < arr[j]) {
                next = arr[j];
                break;
            }
        }
        printf("%d -- %dn", arr[i], next);
    }
}

int main()
{
    int arr[] = { 11, 13, 21, 3 };
    int n = sizeof(arr) / sizeof(arr[0]);
    printNGE(arr, n);
    return 0;
}

Java

// Simple Java program to print next
// greater elements in a given array

class Main {
    /* prints element and NGE pair for
     all elements of arr[] of size n */
    static void printNGE(int arr[], int n)
    {
        int next, i, j;
        for (i = 0; i < n; i++) {
            next = -1;
            for (j = i + 1; j < n; j++) {
                if (arr[i] < arr[j]) {
                    next = arr[j];
                    break;
                }
            }
            System.out.println(arr[i] + " -- " + next);
        }
    }

    public static void main(String args[])
    {
        int arr[] = { 11, 13, 21, 3 };
        int n = arr.length;
        printNGE(arr, n);
    }
}

Python

# Function to print element and NGE pair for all elements of list
def printNGE(arr):

    for i in range(0, len(arr), 1):

        next = -1
        for j in range(i+1, len(arr), 1):
            if arr[i] < arr[j]:
                next = arr[j]
                break

        print(str(arr[i]) + " -- " + str(next))


# Driver program to test above function
arr = [11, 13, 21, 3]
printNGE(arr)

# This code is contributed by Sunny Karira

C#

// Simple C# program to print next
// greater elements in a given array
using System;

class GFG {

    /* prints element and NGE pair for
    all elements of arr[] of size n */
    static void printNGE(int[] arr, int n)
    {
        int next, i, j;
        for (i = 0; i < n; i++) {
            next = -1;
            for (j = i + 1; j < n; j++) {
                if (arr[i] < arr[j]) {
                    next = arr[j];
                    break;
                }
            }
            Console.WriteLine(arr[i] + " -- " + next);
        }
    }

    // driver code
    public static void Main()
    {
        int[] arr = { 11, 13, 21, 3 };
        int n = arr.Length;

        printNGE(arr, n);
    }
}

// This code is contributed by Sam007

PHP

<?php
// Simple PHP program to print next
// greater elements in a given array

/* prints element and NGE pair for
   all elements of arr[] of size n */
function printNGE($arr, $n)
{
    for ($i = 0; $i < $n; $i++)
    {
        $next = -1;
        for ($j = $i + 1; $j < $n; $j++)
        {
            if ($arr[$i] < $arr[$j])
            {
                $next = $arr[$j];
                break;
            }
        }
        echo $arr[$i]." -- ". $next."\n";

    }
}

    // Driver Code
    $arr= array(11, 13, 21, 3);
    $n = count($arr);
    printNGE($arr, $n);

// This code is contributed by Sam007
?>

Javascript

<script>
      // Simple JavaScript program to print
      // next greater elements in a
      // given array

      /* prints element and NGE pair
      for all elements of arr[] of size n */
      function printNGE(arr, n)
      {
        var next, i, j;
        for (i = 0; i < n; i++)
        {
          next = -1;
          for (j = i + 1; j < n; j++)
          {
            if (arr[i] < arr[j])
            {
              next = arr[j];
              break;
            }
          }
          document.write(arr[i] + " -- " + next);
          document.write("<br>");
        }
      }

      // Driver Code
      var arr = [11, 13, 21, 3];
      var n = arr.length;
      printNGE(arr, n);

      // This code is contributed by rdtank.
    </script>

Output

11 --> 13
13 --> 21
21 --> -1
3 --> -1

Time Complexity: O(N²) 
Auxiliary Space: O(1)

Find Next Greater Element using Stack:

The idea is to store the elements for which we have to find the next greater element in a stack and while traversing the array, if we find a greater element, we will pair it with the elements from the stack till the top element of the stack is less than the current element.

illustration:
Below is the illustration of the above approach: 

Follow the steps mentioned below to implement the idea:

• Push the first element to stack.
• Pick the rest of the elements one by one and follow the following steps in the loop. 
  • Mark the current element as next.
  • If the stack is not empty, compare top most element of stack with next.
  • If next is greater than the top element, Pop element from the stack. next is the next greater element for the popped element.
  • Keep popping from the stack while the popped element is smaller than next. next becomes the next greater element for all such popped elements.
• Finally, push the next in the stack.
• After the loop in step 2 is over, pop all the elements from the stack and print -1 as the next element for them.           

Below is the implementation of the above approach:

C++

// A Stack based C++ program to find next
// greater element for all array elements.
#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;

    /* push the first element to stack */
    s.push(arr[0]);

    // iterate for rest of the elements
    for (int i = 1; i < n; i++) {

        if (s.empty()) {
            s.push(arr[i]);
            continue;
        }

        /* if stack is not empty, then
           pop an element from stack.
           If the popped element is smaller
           than next, then
        a) print the pair
        b) keep popping while elements are
        smaller and stack is not empty */
        while (s.empty() == false && s.top() < arr[i]) {
            cout << s.top() << " --> " << arr[i] << endl;
            s.pop();
        }

        /* push next to stack so that we can find
        next greater for it */
        s.push(arr[i]);
    }

    /* After iterating over the loop, the remaining
    elements in stack do not have the next greater
    element, so print -1 for them */
    while (s.empty() == false) {
        cout << s.top() << " --> " << -1 << endl;
        s.pop();
    }
}

/* Driver code */
int main()
{
    int arr[] = { 11, 13, 21, 3 };
    int n = sizeof(arr) / sizeof(arr[0]);
    printNGE(arr, n);
    return 0;
}

C

// A Stack based C program to find next
//  greater element for all array elements.
#include <stdbool.h>
#include <stdio.h>
#include <stdlib.h>
#define STACKSIZE 100

// stack structure
struct stack {
    int top;
    int items[STACKSIZE];
};

// Stack Functions to be used by printNGE()
void push(struct stack* ps, int x)
{
    if (ps->top == STACKSIZE - 1) {
        printf("Error: stack overflown");
        getchar();
        exit(0);
    }
    else {
        ps->top += 1;
        int top = ps->top;
        ps->items[top] = x;
    }
}

bool isEmpty(struct stack* ps)
{
    return (ps->top == -1) ? true : false;
}

int pop(struct stack* ps)
{
    int temp;
    if (ps->top == -1) {
        printf("Error: stack underflow n");
        getchar();
        exit(0);
    }
    else {
        int top = ps->top;
        temp = ps->items[top];
        ps->top -= 1;
        return temp;
    }
}

/* prints element and NGE pair for all elements of
arr[] of size n */
void printNGE(int arr[], int n)
{
    int i = 0;
    struct stack s;
    s.top = -1;
    int element, next;

    /* push the first element to stack */
    push(&s, arr[0]);

    // iterate for rest of the elements
    for (i = 1; i < n; i++) {
        next = arr[i];

        if (isEmpty(&s) == false) {
            // if stack is not empty, then pop an element
            // from stack
            element = pop(&s);

            /* If the popped element is smaller than next,
               then a) print the pair b) keep popping while
               elements are smaller and stack is not empty
             */
            while (element < next) {
                printf("n %d --> %d", element, next);
                if (isEmpty(&s) == true)
                    break;
                element = pop(&s);
            }

            /* If element is greater than next, then push
               the element back */
            if (element > next)
                push(&s, element);
        }

        /* push next to stack so that we can find
           next greater for it */
        push(&s, next);
    }

    /* After iterating over the loop, the remaining
       elements in stack do not have the next greater
       element, so print -1 for them */
    while (isEmpty(&s) == false) {
        element = pop(&s);
        next = -1;
        printf("n %d --> %d", element, next);
    }
}

/* Driver code */
int main()
{
    int arr[] = { 11, 13, 21, 3 };
    int n = sizeof(arr) / sizeof(arr[0]);
    printNGE(arr, n);
    getchar();
    return 0;
}

Java

// Java program to print next
// greater element using stack

public class NGE {
    static class stack {
        int top;
        int items[] = new int[100];

        // Stack functions to be used by printNGE
        void push(int x)
        {
            if (top == 99) {
                System.out.println("Stack full");
            }
            else {
                items[++top] = x;
            }
        }

        int pop()
        {
            if (top == -1) {
                System.out.println("Underflow error");
                return -1;
            }
            else {
                int element = items[top];
                top--;
                return element;
            }
        }

        boolean isEmpty()
        {
            return (top == -1) ? true : false;
        }
    }

    /* prints element and NGE pair for
       all elements of arr[] of size n */
    static void printNGE(int arr[], int n)
    {
        int i = 0;
        stack s = new stack();
        s.top = -1;
        int element, next;

        /* push the first element to stack */
        s.push(arr[0]);

        // iterate for rest of the elements
        for (i = 1; i < n; i++) {
            next = arr[i];

            if (s.isEmpty() == false) {

                // if stack is not empty, then
                // pop an element from stack
                element = s.pop();

                /* If the popped element is smaller than
                   next, then a) print the pair b) keep
                   popping while elements are smaller and
                   stack is not empty */
                while (element < next) {
                    System.out.println(element + " --> "
                                       + next);
                    if (s.isEmpty() == true)
                        break;
                    element = s.pop();
                }

                /* If element is greater than next, then
                   push the element back */
                if (element > next)
                    s.push(element);
            }

            /* push next to stack so that we can find next
               greater for it */
            s.push(next);
        }

        /* After iterating over the loop, the remaining
           elements in stack do not have the next greater
           element, so print -1 for them */
        while (s.isEmpty() == false) {
            element = s.pop();
            next = -1;
            System.out.println(element + " -- " + next);
        }
    }

    // Driver Code
    public static void main(String[] args)
    {
        int arr[] = { 11, 13, 21, 3 };
        int n = arr.length;
        printNGE(arr, n);
    }
}

// Thanks to Rishabh Mahrsee for contributing this code

Python

# Python program to print next greater element using stack

# Stack Functions to be used by printNGE()


def createStack():
    stack = []
    return stack


def isEmpty(stack):
    return len(stack) == 0


def push(stack, x):
    stack.append(x)


def pop(stack):
    if isEmpty(stack):
        print("Error : stack underflow")
    else:
        return stack.pop()


'''prints element and NGE pair for all elements of
   arr[] '''


def printNGE(arr):
    s = createStack()
    element = 0
    next = 0

    # push the first element to stack
    push(s, arr[0])

    # iterate for rest of the elements
    for i in range(1, len(arr), 1):
        next = arr[i]

        if isEmpty(s) == False:

            # if stack is not empty, then pop an element from stack
            element = pop(s)

            '''If the popped element is smaller than next, then
                a) print the pair
                b) keep popping while elements are smaller and
                   stack is not empty '''
            while element < next:
                print(str(element) + " -- " + str(next))
                if isEmpty(s) == True:
                    break
                element = pop(s)

            '''If element is greater than next, then push
               the element back '''
            if element > next:
                push(s, element)

        '''push next to stack so that we can find
           next greater for it '''
        push(s, next)

    '''After iterating over the loop, the remaining
       elements in stack do not have the next greater
       element, so print -1 for them '''

    while isEmpty(s) == False:
        element = pop(s)
        next = -1
        print(str(element) + " -- " + str(next))


# Driver code
arr = [11, 13, 21, 3]
printNGE(arr)

# This code is contributed by Sunny Karira

C#

using System;

// c# program to print next
// greater element using stack

public class NGE {
    public class stack {
        public int top;
        public int[] items = new int[100];

        // Stack functions to be used by printNGE
        public virtual void push(int x)
        {
            if (top == 99) {
                Console.WriteLine("Stack full");
            }
            else {
                items[++top] = x;
            }
        }

        public virtual int pop()
        {
            if (top == -1) {
                Console.WriteLine("Underflow error");
                return -1;
            }
            else {
                int element = items[top];
                top--;
                return element;
            }
        }

        public virtual bool Empty
        {
            get { return (top == -1) ? true : false; }
        }
    }

    /* prints element and NGE pair for
       all elements of arr[] of size n */
    public static void printNGE(int[] arr, int n)
    {
        int i = 0;
        stack s = new stack();
        s.top = -1;
        int element, next;

        /* push the first element to stack */
        s.push(arr[0]);

        // iterate for rest of the elements
        for (i = 1; i < n; i++) {
            next = arr[i];

            if (s.Empty == false) {

                // if stack is not empty, then
                // pop an element from stack
                element = s.pop();

                /* If the popped element is smaller than
                   next, then a) print the pair b) keep
                   popping while elements are smaller and
                   stack is not empty */
                while (element < next) {
                    Console.WriteLine(element + " --> "
                                      + next);
                    if (s.Empty == true) {
                        break;
                    }
                    element = s.pop();
                }

                /* If element is greater than next, then
                   push the element back */
                if (element > next) {
                    s.push(element);
                }
            }

            /* push next to stack so that we can find next
               greater for it */
            s.push(next);
        }

        /* After iterating over the loop, the remaining
           elements in stack do not have the next greater
           element, so print -1 for them */
        while (s.Empty == false) {
            element = s.pop();
            next = -1;
            Console.WriteLine(element + " -- " + next);
        }
    }

    // Driver Code
    public static void Main(string[] args)
    {
        int[] arr = new int[] { 11, 13, 21, 3 };
        int n = arr.Length;
        printNGE(arr, n);
    }
}

// This code is contributed by Shrikant13

Javascript

<script>

// A Stack based Javascript program to find next
// greater element for all array elements.

/* prints element and NGE pair for all
elements of arr[] of size n */
function printNGE(arr, n)
{
    var s = [];

    /* push the first element to stack */
    s.push(arr[0]);

    // iterate for rest of the elements
    for (var i = 1; i < n; i++)
    {

        if (s.length == 0) {
            s.push(arr[i]);
            continue;
        }

        /* if stack is not empty, then
           pop an element from stack.
           If the popped element is smaller
           than next, then
        a) print the pair
        b) keep popping while elements are
        smaller and stack is not empty */
        while (s.length ==0 == false
               && s[s.length-1] < arr[i])
        {
            document.write( s[s.length-1]
                 + " --> " + arr[i]+"<br>");
            s.pop();
        }

        /* push next to stack so that we can find
        next greater for it */
        s.push(arr[i]);
    }

    /* After iterating over the loop, the remaining
    elements in stack do not have the next greater
    element, so print -1 for them */
    while (s.length !=0) {
        document.write( s[s.length-1] + " --> " + -1+ "<br>" );
        s.pop();
    }
}

/* Driver code */
var arr = [11, 13, 21, 3];
var n = arr.length;
printNGE(arr, n);

</script>

Output

11 --> 13
13 --> 21
3 --> -1
21 --> -1

Time Complexity: O(N) 
Auxiliary Space: O(N) 

Find Next Greater Element using Map: 

In this particular approach we are using the map as our main stack

• This is same as above method but the elements are pushed and popped only once into the stack. The array is changed in place. The array elements are pushed into the stack until it finds a greatest element in the right of array. In other words the elements are popped from stack when top of the stack value is smaller in the current array element.
• Once all the elements are processed in the array but stack is not empty. The left out elements in the stack doesn’t encounter any greatest element . So pop the element from stack and change it’s index value as -1 in the array.

C++

// C++ code to implement the approach

#include <bits/stdc++.h>

using namespace std;

void nextLargerElement(int arr[], int n)
{
    vector<unordered_map<string, int> > s;

    // iterating over the array
    for (int i = 0; i < n; i++) {
        while (s.size() > 0
               && s[s.size() - 1]["value"] < arr[i]) {
            // updating the array as per the stack top
            unordered_map<string, int> d = s[s.size() - 1];
            s.pop_back();
            arr[d["ind"]] = arr[i];
        }
        // pushing values to stack
        unordered_map<string, int> e;

        e["value"] = arr[i];
        e["ind"] = i;
        s.push_back(e);
    }

    // updating the array as per the stack top
    while (s.size() > 0) {
        unordered_map<string, int> d = s[s.size() - 1];
        s.pop_back();
        arr[d["ind"]] = -1;
    }
}

// Driver Code

int main()
{
    int arr[] = { 6, 8, 0, 1, 3 };
    int n = 5;

    // Function call
    nextLargerElement(arr, n);
    for (int i = 0; i < n; i++)
        cout << arr[i] << " ";
}

// This code is contributed by phasing17

Java

// Java code to implement the approach

import java.util.*;

class GFG
{
    static void nextLargerElement(int[] arr, int n)
    {
        ArrayList<HashMap<String, Integer> > s = new  ArrayList<HashMap<String, Integer> >();

        // iterating over the array
        for (int i = 0; i < n; i++) {
            while (s.size() > 0
                   && s.get(s.size() - 1).get("value") < arr[i]) {
                // updating the array as per the stack top
                HashMap<String, Integer> d = s.get(s.size() - 1);
                s.remove(s.size() - 1);
                arr[d.get("ind")] = arr[i];
            }
            // pushing values to stack
            HashMap<String, Integer> e = new HashMap<String, Integer>();

            e.put("value", arr[i]);
            e.put("ind", i);
            s.add(e);
        }

        // updating the array as per the stack top
        while (s.size() > 0) {
            HashMap<String, Integer> d = s.get(s.size() - 1);
            s.remove(s.size() - 1);
            arr[d.get("ind")] = -1;
        }
    }

    // Driver Code

    public static void main(String[] args)
    {
        int[] arr = { 6, 8, 0, 1, 3 };
        int n = 5;

        // Function call
        nextLargerElement(arr, n);
        for (int i = 0; i < n; i++)
            System.out.print(arr[i] + " ");
    }
}

// This code is contributed by phasing17

Python3

# Python3 code
class Solution:
    def nextLargerElement(self, arr, n):
        # code here
        s = []
        for i in range(len(arr)):
            while s and s[-1].get("value") < arr[i]:
                d = s.pop()
                arr[d.get("ind")] = arr[i]
            s.append({"value": arr[i], "ind": i})
        while s:
            d = s.pop()
            arr[d.get("ind")] = -1
        return arr


if __name__ == "__main__":
    print(Solution().nextLargerElement([6, 8, 0, 1, 3], 5))

C#

// C# code to implement the approach

using System;
using System.Collections.Generic;

class GFG
{
    static void nextLargerElement(int[] arr, int n)
    {
        List<Dictionary<string, int> > s = new List<Dictionary<string, int> >();

        // iterating over the array
        for (int i = 0; i < n; i++) {
            while (s.Count > 0
                   && s[s.Count - 1]["value"] < arr[i]) {
                // updating the array as per the stack top
                Dictionary<string, int> d = s[s.Count - 1];
                s.RemoveAt(s.Count - 1);
                arr[d["ind"]] = arr[i];
            }
            // pushing values to stack
            Dictionary<string, int> e = new Dictionary<string, int>();

            e["value"] = arr[i];
            e["ind"] = i;
            s.Add(e);
        }

        // updating the array as per the stack top
        while (s.Count > 0) {
            Dictionary<string, int> d = s[s.Count - 1];
            s.RemoveAt(s.Count - 1);
            arr[d["ind"]] = -1;
        }
    }

    // Driver Code

    public static void Main(string[] args)
    {
        int[] arr = { 6, 8, 0, 1, 3 };
        int n = 5;

        // Function call
        nextLargerElement(arr, n);
        for (int i = 0; i < n; i++)
            Console.Write(arr[i] + " ");
    }
}

// This code is contributed by phasing17

Javascript

// JS code to implement the approach
function nextLargerElement(arr, n)
{
    var s = [];

    // iterating over the array
    for (var i = 0; i < arr.length; i++)
    {
        while (s.length > 0 && s[s.length - 1]["value"] < arr[i])
        {
            // updating the array as per the stack top
            var d = s.pop();
            arr[d["ind"]] = arr[i];
        }
        // pushing values to stack
        s.push({"value": arr[i], "ind": i});
    }

    // updating the array as per the stack top
    while (s.length > 0)
    {
        d = s.pop();
        arr[d["ind"]] = -1;
    }
    return arr;

}

// Driver Code
var arr = [6, 8, 0, 1, 3];
var n = 5;

// Function call
console.log(nextLargerElement(arr, n));

// This code is contributed by phasing17

Output

8 -1 1 3 -1 

Time Complexity: O(N) 
Auxiliary Space: O(N)

Next Greater Element in O(n) Time Complexity and O(1) Space Complexity

In this particular approach we follows these steps:

1. Initialize an output array v with -1 as the initial value for each element.
2. Set the maximum element seen so far (mx) to be the last element of the input array.
3. Traverse the input array from right to left.
4. For each element arr[i], if the next element arr[i+1] is greater than arr[i], then set v[i] to arr[i+1] since arr[i+1] is the next greater element for arr[i].
5. If arr[i+1] is not greater than arr[i], then check v[i+1]. If v[i+1] is greater than arr[i], then set v[i] to v[i+1] since v[i+1] is the next greater element for arr[i].
6. If v[i+1] is not greater than arr[i], then check the maximum element seen so far (mx). If mx is greater than arr[i], then traverse the subarray to the right of arr[i] until an element greater than arr[i] is found. Set v[i] to that element. If no such element is found, set v[i] to -1.
7. If mx is not greater than arr[i], set v[i] to -1.
8. Update the maximum element seen so far (mx) to be the maximum of arr[i] and mx.
9. Return the output array v.

C++

#include <iostream>
#include <vector>
using namespace std;

class Solution {
public:
    vector<long long>
    nextLargerElement(vector<long long> arr, int n)
    {
        // Initialize a vector to store the next greater
        // elements.
        vector<long long> v(n, -1);

        // Initialize a variable to keep track of the
        // maximum element seen so far.
        long long mx = arr[n - 1];

        // Iterate over the array from right to left.
        for (int i = n - 2; i >= 0; i--) {
            // If the element to the right is greater than
            // the current element, then the next greater
            // element for the current element is the
            // element to the right.
            if (arr[i + 1] > arr[i]) {
                v[i] = arr[i + 1];
            }
            else {
                // If the next greater element for the
                // element to the right is greater than the
                // current element, then the next greater
                // element for the current element is the
                // next greater element for the element to
                // the right.
                if (v[i + 1] > arr[i]) {
                    v[i] = v[i + 1];
                }
                else if (mx > arr[i]) {
                    // If there is no next greater element
                    // to the right that is greater than the
                    // current element, then we need to find
                    // the next greater element that is
                    // greater than the current element and
                    // to its right. Initialize a variable
                    // to keep track of the index of the
                    // next greater element.
                    int k = i + 1;
                    // Iterate over the array from the
                    // current element to its right to find
                    // the next greater element.
                    while (arr[k] <= arr[i]) {
                        k++;
                    }
                    v[i] = arr[k];
                }
                else {
                    // If there is no next greater element
                    // to the right that is greater than the
                    // current element and there is no next
                    // greater element to the right that is
                    // greater than the next greater element
                    // for the current element, then there
                    // is no next greater element for the
                    // current element.
                    v[i] = -1;
                }
            }
            // Update the maximum element seen so far.
            mx = max(arr[i], mx);
        }
        // Return the vector containing the next greater
        // elements.
        return v;
    }
};

int main()
{
    // Input array
    vector<long long> arr = { 3, 8, 4, 1, 2, 6, 7, 2 };
    int n = arr.size();
    // Create an instance of the Solution class
    Solution sol;
    // Call the nextLargerElement function
    vector<long long> result
        = sol.nextLargerElement(arr, n);
    // Print the result
    cout << "For the input array [3, 8, 4, 1, 2, 6, 7, 2], "
            "the next greater elements are: ";
    for (int i = 0; i < n; i++) {
        cout << result[i] << " ";
    }
    cout << endl;
    return 0;
}

Java

import java.util.*;

class Solution {
    public List<Long> nextLargerElement(List<Long> arr,
                                        int n)
    {
        // Initialize a list to store the next greater
        // elements.
        List<Long> v
            = new ArrayList<>(Collections.nCopies(n, -1L));

        // Initialize a variable to keep track of the
        // maximum element seen so far.
        long mx = arr.get(n - 1);

        // Iterate over the array from right to left.
        for (int i = n - 2; i >= 0; i--) {
            // If the element to the right is greater than
            // the current element, then the next greater
            // element for the current element is the
            // element to the right.
            if (arr.get(i + 1) > arr.get(i)) {
                v.set(i, arr.get(i + 1));
            }
            else {
                // If the next greater element for the
                // element to the right is greater than the
                // current element, then the next greater
                // element for the current element is the
                // next greater element for the element to
                // the right.
                if (v.get(i + 1) > arr.get(i)) {
                    v.set(i, v.get(i + 1));
                }
                else if (mx > arr.get(i)) {
                    // If there is no next greater element
                    // to the right that is greater than the
                    // current element, then we need to find
                    // the next greater element that is
                    // greater than the current element and
                    // to its right. Initialize a variable
                    // to keep track of the index of the
                    // next greater element.
                    int k = i + 1;
                    // Iterate over the array from the
                    // current element to its right to find
                    // the next greater element.
                    while (arr.get(k) <= arr.get(i)) {
                        k++;
                    }
                    v.set(i, arr.get(k));
                }
                else {
                    // If there is no next greater element
                    // to the right that is greater than the
                    // current element and there is no next
                    // greater element to the right that is
                    // greater than the next greater element
                    // for the current element, then there
                    // is no next greater element for the
                    // current element.
                    v.set(i, -1L);
                }
            }
            // Update the maximum element seen so far.
            mx = Math.max(arr.get(i), mx);
        }
        // Return the list containing the next greater
        // elements.
        return v;
    }
}

public class Main {
    public static void main(String[] args)
    {
        // Input array
        List<Long> arr
            = Arrays.asList(3L, 8L, 4L, 1L, 2L, 6L, 7L, 2L);
        int n = arr.size();
        // Create an instance of the Solution class
        Solution sol = new Solution();
        // Call the nextLargerElement function
        List<Long> result = sol.nextLargerElement(arr, n);
        // Print the result
        System.out.print(
            "For the input array [3, 8, 4, 1, 2, 6, 7, 2], the next greater elements are: ");
        for (int i = 0; i < n; i++) {
            System.out.print(result.get(i) + " ");
        }
        System.out.println();
    }
}

Python

class Solution:
    def nextLargerElement(self, arr, n):
        # Initialize a list to store the next greater elements.
        v = [-1] * n

        # Initialize a variable to keep track of the maximum element seen so far.
        mx = arr[n-1]

        # Iterate over the array from right to left.
        for i in range(n-2, -1, -1):
            # If the element to the right is greater than the current element,
            # then the next greater element for the current element is the element to the right.
            if arr[i+1] > arr[i]:
                v[i] = arr[i+1]
            else:
                # If the next greater element for the element to the right is greater than the current element,
                # then the next greater element for the current element is the next greater element for the element to the right.
                if v[i+1] > arr[i]:
                    v[i] = v[i+1]
                elif mx > arr[i]:
                    # If there is no next greater element to the right that is greater than the current element,
                    # then we need to find the next greater element that is greater than the current element and to its right.
                    # Initialize a variable to keep track of the index of the next greater element.
                    k = i+1
                    # Iterate over the array from the current element to its right to find the next greater element.
                    while arr[k] <= arr[i]:
                        k += 1
                    v[i] = arr[k]
                else:
                    # If there is no next greater element to the right that is greater than the current element
                    # and there is no next greater element to the right that is greater than the next greater element for the current element,
                    # then there is no next greater element for the current element.
                    v[i] = -1
            # Update the maximum element seen so far.
            mx = max(arr[i], mx)

        # Return the list containing the next greater elements.
        return v


# Input array
arr = [3, 8, 4, 1, 2, 6, 7, 2]
n = len(arr)

# Create an instance of the Solution class
sol = Solution()

# Call the nextLargerElement function
result = sol.nextLargerElement(arr, n)

# Print the result
print "For the input array [3, 8, 4, 1, 2, 6, 7, 2], the next greater elements are: ",
for i in range(n):
    print result[i],
print

C#

using System;
using System.Collections.Generic;

class Solution
{
  public List<long> nextLargerElement(List<long> arr, int n)
  {
    // Initialize a list to store the next greater elements.
    List<long> v = new List<long>(n);
    for (int i = 0; i < n; i++) {
      v.Add(-1);
    }

    // Initialize a variable to keep track of the maximum element seen so far.
    long mx = arr[n - 1];

    // Iterate over the array from right to left.
    for (int i = n - 2; i >= 0; i--) {
      // If the element to the right is greater than the current element, then the next greater element for the current element is the element to the right.
      if (arr[i + 1] > arr[i]) {
        v[i] = arr[i + 1];
      }
      else {
        // If the next greater element for the element to the right is greater than the current element, then the next greater element for the current element is the next greater element for the element to the right.
        if (v[i + 1] > arr[i]) {
          v[i] = v[i + 1];
        }
        else if (mx > arr[i]) {
          // If there is no next greater element to the right that is greater than the current element, then we need to find the next greater element that is greater than the current element and to its right. Initialize a variable to keep track of the index of the next greater element.
          int k = i + 1;
          // Iterate over the array from the current element to its right to find the next greater element.
          while (arr[k] <= arr[i]) {
            k++;
          }
          v[i] = arr[k];
        }
        else {
          // If there is no next greater element to the right that is greater than the current element and there is no next greater element to the right that is greater than the next greater element for the current element, then there is no next greater element for the current element.
          v[i] = -1;
        }
      }
      // Update the maximum element seen so far.
      mx = Math.Max(arr[i], mx);
    }
    // Return the list containing the next greater elements.
    return v;
  }
}

class Gfg
{
  static void Main(string[] args)
  {
    // Input list
    List<long> arr = new List<long> { 3, 8, 4, 1, 2, 6, 7, 2 };
    int n = arr.Count;
    // Create an instance of the Solution class
    Solution sol = new Solution();
    // Call the nextLargerElement function
    List<long> result = sol.nextLargerElement(arr, n);
    // Print the result
    Console.Write("For the input array [3, 8, 4, 1, 2, 6, 7, 2], the next greater elements are: ");
    for (int i = 0; i < n; i++) {
      Console.Write(result[i] + " ");
    }
    Console.WriteLine();
  }
}

Javascript

// Define a Solution class with a function to find the next greater element
class Solution {
    nextLargerElement(arr, n) {
        // Initialize a vector to store the next greater elements
        let v = new Array(n).fill(-1);

        // Initialize a variable to keep track of the maximum element seen so far
        let mx = arr[n - 1];

        // Iterate over the array from right to left
        for (let i = n - 2; i >= 0; i--) {
            // If the element to the right is greater than the current element,
            // then the next greater element for the current element is the element to the right.
            if (arr[i + 1] > arr[i]) {
                v[i] = arr[i + 1];
            } else {
                // If the next greater element for the element to the right is greater than the
                // current element, then the next greater element for the current element is the
                // next greater element for the element to the right.
                if (v[i + 1] > arr[i]) {
                    v[i] = v[i + 1];
                } else if (mx > arr[i]) {
                    // If there is no next greater element to the right that is greater than the
                    // current element, then we need to find the next greater element that is
                    // greater than the current element and to its right. Initialize a variable
                    // to keep track of the index of the next greater element.
                    let k = i + 1;
                    // Iterate over the array from the current element to its right to find the
                    // next greater element.
                    while (arr[k] <= arr[i]) {
                        k++;
                    }
                    v[i] = arr[k];
                } else {
                    // If there is no next greater element to the right that is greater than the
                    // current element and there is no next greater element to the right that is
                    // greater than the next greater element for the current element, then there
                    // is no next greater element for the current element.
                    v[i] = -1;
                }
            }
            // Update the maximum element seen so far.
            mx = Math.max(arr[i], mx);
        }
        // Return the vector containing the next greater elements.
        return v;
    }
}

// Define an input array
let arr = [3, 8, 4, 1, 2, 6, 7, 2];
let n = arr.length;

// Create an instance of the Solution class
let sol = new Solution();

// Call the nextLargerElement function
let result = sol.nextLargerElement(arr, n);

// Print the result
console.log("For the input array [3, 8, 4, 1, 2, 6, 7, 2], the next greater elements are: ");
console.log(result.join(" "));

Output

For the input array [3, 8, 4, 1, 2, 6, 7, 2], the next greater elements are: 8 -1 6 2 6 7 -1 -1 

Time complexity: O(N), Where N is the size of the array
Auxiliary Space: O(1)