Next Greater Element in Array (original) (raw)
Last Updated : 10 Sep, 2025
Given an array arr[] of integers, determine the **Next Greater Element (NGE) for every element in the array, maintaining the order of appearance.
- The Next Greater Element for an element x is defined as the first element to the right of x in the array that is strictly greater than x.
- If no such element exists for an element, its Next Greater Element is -1.
**Examples:
**Input: arr[] = [1, 3, 2, 4]
**Output: [3, 4, 4, -1]
**Explanation: The next larger element to 1 is 3, 3 is 4, 2 is 4 and for 4, since it doesn't exist, it is -1.**Input: arr[] = [6, 8, 0, 1, 3]
**Output: [8, -1, 1, 3, -1]
**Explanation: The next larger element to 6 is 8, for 8 there is no larger elements hence it is -1, for 0 it is 1 , for 1 it is 3 and then for 3 there is no larger element on right and hence -1.
Table of Content
- [Naive Approach] Using Nested Loops - O(n^2) Time and O(1) Space
- [Expected Approach] Using Stack - O(n) Time and O(n) Space
[Naive Approach] Using Nested Loops - O(n2) Time and O(1) Space
The idea is to look at all the elements to its right and check if there's a larger element for each element in the array. If we find one, we store it as the next greater element. If no greater element is found, we store -1. This is done using two nested loops: the outer loop goes through each element, and the inner loop searches the rest of the array for a greater element.
C++ `
#include #include using namespace std;
vector nextLargerElement(vector &arr) { int n = arr.size(); vector res(n, -1);
// iterate through each element in the array
for (int i = 0; i < n; i++) {
// check for the next greater element
// in the rest of the array
for (int j = i + 1; j < n; j++) {
if (arr[j] > arr[i]) {
res[i] = arr[j];
break;
}
}
}
return res;}
int main() {
vector<int> arr = {6, 8, 0, 1, 3};
vector<int> res = nextLargerElement(arr);
for (int x : res) {
cout << x << " ";
}
return 0;}
Java
import java.util.ArrayList;
class GfG {
static ArrayList<Integer> nextLargerElement(int[] arr) {
int n = arr.length;
ArrayList<Integer> res = new ArrayList<>();
// initialize res with -1 for all elements
for (int i = 0; i < n; i++) {
res.add(-1);
}
// iterate through each element in the array
for (int i = 0; i < n; i++) {
// check for the next greater element
// in the rest of the array
for (int j = i + 1; j < n; j++) {
if (arr[j] > arr[i]) {
res.set(i, arr[j]);
break;
}
}
}
return res;
}
public static void main(String[] args) {
int[] arr = { 6, 8, 0, 1, 3 };
ArrayList<Integer> res = nextLargerElement(arr);
for (int x : res) {
System.out.print(x + " ");
}
}}
Python
def nextLargerElement(arr): n = len(arr) res = [-1] * n
# Iterate through each element in the array
for i in range(n):
# Check for the next greater element
# in the rest of the array
for j in range(i + 1, n):
if arr[j] > arr[i]:
res[i] = arr[j]
break
return resif name == "main": arr = [6, 8, 0, 1, 3]
res = nextLargerElement(arr)
print(" ".join(map(str, res)))C#
using System; using System.Collections.Generic;
class GfG {
static List<int> nextLargerElement(int[] arr) {
int n = arr.Length;
List<int> res = new List<int>(new int[n]);
// Initialize res with -1 for all elements
for (int i = 0; i < n; i++) {
res[i] = -1;
}
// Iterate through each element in the array
for (int i = 0; i < n; i++) {
// Check for the next greater element
// in the rest of the array
for (int j = i + 1; j < n; j++) {
if (arr[j] > arr[i]) {
res[i] = arr[j];
break;
}
}
}
return res;
}
static void Main(string[] args) {
int[] arr = { 6, 8, 0, 1, 3 };
List<int> res = nextLargerElement(arr);
foreach(int x in res) { Console.Write(x + " "); }
}}
JavaScript
function nextLargerElement(arr) {
let n = arr.length;
let res = new Array(n).fill(-1);
// Iterate through each element in the array
for (let i = 0; i < n; i++) {
// Check for the next greater element
// in the rest of the array
for (let j = i + 1; j < n; j++) {
if (arr[j] > arr[i]) {
res[i] = arr[j];
break;
}
}
}
return res;}
// Driver Code let arr = [ 6, 8, 0, 1, 3 ]; let res = nextLargerElement(arr); console.log(res.join(" "));
`
[Expected Approach] Using Stack - O(n) Time and O(n) Space
The idea is to use a monotonic decreasing stack (stack that maintains elements in decreasing order). We traverse the array from right to left. For each element, we pop elements from the stack that are smaller than or equal to it, since they cannot be the next greater element. If the stack is not empty, the top of the stack is the next greater element. Finally, we push the current element onto the stack.
**Illustration:
C++ `
#include #include #include using namespace std;
vector nextLargerElement(vector &arr) { int n = arr.size(); vector res(n, -1); stack stk;
for (int i = n - 1; i >= 0; i--) {
// Pop elements from the stack that are less
// than or equal to the current element
while (!stk.empty() && stk.top() <= arr[i]) {
stk.pop();
}
// If the stack is not empty, the top element
// is the next greater element
if (!stk.empty()) {
res[i] = stk.top();
}
// Push the current element onto the stack
stk.push(arr[i]);
}
return res;}
int main() {
vector<int> arr = {6, 8, 0, 1, 3};
vector<int> res = nextLargerElement(arr);
for (int x : res) {
cout << x << " ";
}
return 0;}
Java
import java.util.ArrayList; import java.util.Stack;
class GfG {
static ArrayList<Integer> nextLargerElement(int[] arr) {
int n = arr.length;
ArrayList<Integer> res = new ArrayList<>();
Stack<Integer> stk = new Stack<>();
// Initialize res with -1 for all elements
for (int i = 0; i < n; i++) {
res.add(-1);
}
for (int i = n - 1; i >= 0; i--) {
// Pop elements from the stack that are less
// than or equal to the current element
while (!stk.isEmpty() && stk.peek() <= arr[i]) {
stk.pop();
}
// If the stack is not empty, the top element
// is the next greater element
if (!stk.isEmpty()) {
res.set(i, stk.peek());
}
// Push the current element onto the stack
stk.push(arr[i]);
}
return res;
}
public static void main(String[] args) {
int[] arr = { 6, 8, 0, 1, 3 };
ArrayList<Integer> res = nextLargerElement(arr);
for (int x : res) {
System.out.print(x + " ");
}
}}
Python
def nextLargerElement(arr):
n = len(arr)
res = [-1] * n
stk = []
# Traverse the array from right to left
for i in range(n - 1, -1, -1):
# Pop elements from the stack that are less
# than or equal to the current element
while stk and arr[stk[-1]] <= arr[i]:
stk.pop()
# If the stack is not empty, the element at the
# top of the stack is the next greater element
if stk:
res[i] = arr[stk[-1]]
# Push the current index onto the stack
stk.append(i)
return resif name == "main":
arr = [6, 8, 0, 1, 3]
res = nextLargerElement(arr)
print(" ".join(map(str, res)))C#
using System; using System.Collections.Generic;
class GfG {
static List<int> nextLargerElement(int[] arr) {
int n = arr.Length;
List<int> res = new List<int>(new int[n]);
Stack<int> stk = new Stack<int>();
// Initialize res with -1 for all elements
for (int i = 0; i < n; i++) {
res[i] = -1;
}
// Traverse the array from right to left
for (int i = n - 1; i >= 0; i--) {
// Pop elements from the stack that are less
// than or equal to the current element
while (stk.Count > 0 && stk.Peek() <= arr[i]) {
stk.Pop();
}
// If the stack is not empty, the top element
// is the next greater element
if (stk.Count > 0) {
res[i] = stk.Peek();
}
// Push the current element onto the stack
stk.Push(arr[i]);
}
return res;
}
static void Main(string[] args) {
int[] arr = { 6, 8, 0, 1, 3 };
List<int> res = nextLargerElement(arr);
foreach (int x in res) {
Console.Write(x + " ");
}
}}
JavaScript
function nextLargerElement(arr) {
let n = arr.length;
let res = new Array(n).fill(-1);
let stk = [];
// Traverse the array from right to left
for (let i = n - 1; i >= 0; i--) {
// Pop elements from the stack that are less
// than or equal to the current element
while (stk.length > 0
&& stk[stk.length - 1] <= arr[i]) {
stk.pop();
}
// If the stack is not empty, the top element
// is the next greater element
if (stk.length > 0) {
res[i] = stk[stk.length - 1];
}
// Push the current element onto the stack
stk.push(arr[i]);
}
return res;}
// Driver Code let arr = [ 6, 8, 0, 1, 3 ]; let res = nextLargerElement(arr); console.log(res.join(" "));
`