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

Last Updated : 12 Feb, 2025
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Given an array arr[] of integers, the task is to find the Next Greater Element for each element of the array in order of their appearance in the array.
Note: 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. 

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.

Input: arr[] = [50, 40, 30, 10]
Output: [-1, -1, -1, -1]
Explanation: There is no greater element for any of the elements in the array, so all are -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(n^2) 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++ 
// C++ implementation to find the next // greater element using two loops #include <iostream> #include <vector> using namespace std;  vector<int> nextLargerElement(vector<int> &arr) {     int n = arr.size();     vector<int> 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 
// Java implementation to find the next // greater element using two loops 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 
# Python implementation to find the next #  greater element using two loops 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 res  if __name__ == "__main__":     arr = [6, 8, 0, 1, 3]      res = nextLargerElement(arr)     print(" ".join(map(str, res)))
C# 
// C# implementation to find the next // greater element using two loops using System;  class GfG {      static int[] nextLargerElement(int[] arr) {         int n = arr.Length;         int[] res = 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 };         int[] res = nextLargerElement(arr);         foreach(int x in res) { Console.Write(x + " "); }     } }
JavaScript 
// Javascript implementation to find the //  next greater element using two loops 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(" "));

Output
8 -1 1 3 -1

[Expected Approach] Using Stack - O(n) Time and O(n) Space

The idea is to use stack to find the next greater element by using the Last-In-First-Out (LIFO) property. We traverse the array from right to left. For each element, we remove elements from the stack that are smaller than or equal to it, as they cannot be the next greater element. If the stack is not empty after this, the top element of the stack is the next greater element for the current element. We then push the current element onto the stack.

Illustration:


C++ 
// C++ implementation to find the next // greater element using a stack #include <iostream> #include <vector> #include <stack> using namespace std;  vector<int> nextLargerElement(vector<int> &arr) {     int n = arr.size();     vector<int> res(n, -1);     stack<int> stk;      // 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.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 
// Java implementation to find the next // greater element using a stack 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);         }          // 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.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 
# Python implementation to find the next # greater element using a stack 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 res  if __name__ == "__main__":        arr = [6, 8, 0, 1, 3]     res = nextLargerElement(arr)     print(" ".join(map(str, res)))
C# 
// C# implementation to find the next // greater element using a stack using System; using System.Collections.Generic;  class GfG {      static int[] nextLargerElement(int[] arr) {         int n = arr.Length;         int[] res = 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 };         int[] res = nextLargerElement(arr);         foreach(int x in res) { Console.Write(x + " "); }     } }
JavaScript 
// JavaScript implementation to find the // next greater element using a stack 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(" "));

Output
8 -1 1 3 -1

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Next Greater Frequency Element

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