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# 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

Recommended Problem

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 ` `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);` `    ``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 `   `/* 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);` `    ``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

 ``

## Javascript

 ``

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 ` `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);`   `    ``// 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);` `    ``printNGE(arr, n);` `    ``return` `0;` `}`

## C

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

 ``

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 `   `using` `namespace` `std;`   `void` `nextLargerElement(``int` `arr[], ``int` `n)` `{` `    ``vector > 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 d = s[s.size() - 1];` `            ``s.pop_back();` `            ``arr[d[``"ind"``]] = arr[i];` `        ``}` `        ``// pushing values to stack` `        ``unordered_map 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 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 > s = ``new`  `ArrayList >();` `    `  `        ``// 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 d = s.get(s.size() - ``1``);` `                ``s.remove(s.size() - ``1``);` `                ``arr[d.get(``"ind"``)] = arr[i];` `            ``}` `            ``// pushing values to stack` `            ``HashMap e = ``new` `HashMap();` `    `  `            ``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 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 > s = ``new` `List >();` `    `  `        ``// 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 ` `#include ` `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 nextLargerElement(List arr,` `                                        ``int` `n)` `    ``{` `        ``// Initialize a list to store the next greater` `        ``// elements.` `        ``List 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 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 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)

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