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Find the next greater element in a Circular Array

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Given a circular array arr[] of N integers such that the last element of the given array is adjacent to the first element of the array, the task is to print the Next Greater Element in this circular array. Elements for which no greater element exist, consider the next greater element as “-1”.

Examples:

Input: arr[] = {5, 6, 7} 
Output: {6, 7, -1} 
Explanation: 
Next Greater Element for 5 is 6, for 6 is 7, and for 7 is -1 as we don’t have any element greater than itself so its -1.

Input: arr[] = {4, -2, 5, 8} 
Output: {5, 5, 8, -1} 
Explanation: 
Next Greater Element for 4 is 5, for -2 its 5, for 5 is 8, and for 8 is -1 as we don’t have any element greater than itself so its -1, and for 3 its 4. 

Approach: This problem can be solved using Greedy Approach. Below are the steps: 

  • For the property of the circular array to be valid append the given array elements to the same array once again.

For Example: 

Let arr[] = {1, 4, 3} 
After appending the same set of elements arr[] becomes 
arr[] = {1, 4, 3, 1, 4, 3} 

  • Find the next greater element till N elements in the above array formed.
  • If any greater element is found then print that element, else print “-1”.

Below is the implementation of the above approach: 

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the NGE
void printNGE(int A[], int n)
{
    // Formation of circular array
    int arr[2 * n];
    // Append the given array element twice
    for (int i = 0; i < 2 * n; i++)
        arr[i] = A[i % n];
    int next, i, j;
    // Iterate for all the elements of the array
    for (i = 0; i < n; i++) {
        // Initialise NGE as -1
        next = -1;
        for (j = i + 1; j < 2 * n; j++) {
            // Checking for next greater element
            if (arr[i] < arr[j]) {
                next = arr[j];
                break;
            }
        }
        // Print the updated NGE
        cout << next << ", ";
    }
}
 
// Driver Code
int main()
{
    // Given array arr[]
    int arr[] = { 1, 2, 1 };
    int N = sizeof(arr) / sizeof(arr[0]);
    // Function call
    printNGE(arr, N);
    return 0;
}
 
// This code is contributed by Aditya Kumar (adityakumar129)


C




// C program for the above approach
 
#include <stdio.h>
 
// Function to find the NGE
void printNGE(int A[], int n)
{
    // Formation of circular array
    int arr[2 * n];
    // Append the given array element twice
    for (int i = 0; i < 2 * n; i++)
        arr[i] = A[i % n];
    int next, i, j;
    // Iterate for all the elements of the array
    for (i = 0; i < n; i++) {
        // Initialise NGE as -1
        next = -1;
        for (j = i + 1; j < 2 * n; j++) {
            // Checking for next greater element
            if (arr[i] < arr[j]) {
                next = arr[j];
                break;
            }
        }
        // Print the updated NGE
        printf("%d, ",next);
    }
}
 
// Driver Code
int main()
{
    // Given array arr[]
    int arr[] = { 1, 2, 1 };
    int N = sizeof(arr) / sizeof(arr[0]);
    // Function call
    printNGE(arr, N);
    return 0;
}
 
// This code is contributed by Aditya Kumar (adityakumar129)


Java




// Java program for the above approach
import java.util.*;
class GFG {
 
    // Function to find the NGE
    static void printNGE(int[] A, int n)
    {
        // Formation of circular array
        int[] arr = new int[2 * n];
        // Append the given array element twice
        for (int i = 0; i < 2 * n; i++)
            arr[i] = A[i % n];
        int next;
        // Iterate for all the elements of the array
        for (int i = 0; i < n; i++) {
            // Initialise NGE as -1
            next = -1;
            for (int j = i + 1; j < 2 * n; j++) {
                // Checking for next greater element
                if (arr[i] < arr[j]) {
                    next = arr[j];
                    break;
                }
            }
            // Print the updated NGE
            System.out.print(next + ", ");
        }
    }
 
    // Driver Code
    public static void main(String args[])
    {
        // Given array arr[]
        int[] arr = { 1, 2, 1 };
        int N = arr.length;
        // Function call
        printNGE(arr, N);
    }
}
 
// This code is contributed by Aditya Kumar (adityakumar129)


Python3




# Python3 program for the above approach
 
# Function to find the NGE
def printNGE(A, n):
 
    # Formation of circular array
    arr = [0] * (2 * n)
 
    # Append the given array
    # element twice
    for i in range(2 * n):
        arr[i] = A[i % n]
 
    # Iterate for all the
    # elements of the array
    for i in range(n):
 
        # Initialise NGE as -1
        next = -1
 
        for j in range(i + 1, 2 * n):
 
            # Checking for next
            # greater element
            if(arr[i] < arr[j]):
                next = arr[j]
                break
 
        # Print the updated NGE
        print(next, end = ", ")
 
# Driver code
if __name__ == '__main__':
 
    # Given array arr[]
    arr = [ 1, 2, 1 ]
 
    N = len(arr)
 
    # Function call
    printNGE(arr, N)
 
# This code is contributed by Shivam Singh


C#




// C# program for the above approach
using System;
class GFG{
 
// Function to find the NGE
static void printNGE(int []A, int n)
{
 
    // Formation of circular array
    int []arr = new int[2 * n];
 
    // Append the given array element twice
    for(int i = 0; i < 2 * n; i++)
       arr[i] = A[i % n];
 
    int next;
 
    // Iterate for all the
    // elements of the array
    for(int i = 0; i < n; i++)
    {
 
       // Initialise NGE as -1
       next = -1;
        
       for(int j = i + 1; j < 2 * n; j++)
       {
            
          // Checking for next
          // greater element
          if (arr[i] < arr[j])
          {
              next = arr[j];
              break;
          }
       }
        
       // Print the updated NGE
       Console.Write(next + ", ");
    }
}
 
// Driver Code
public static void Main()
{
     
    // Given array arr[]
    int []arr = { 1, 2, 1 };
 
    int N = arr.Length;
 
    // Function call
    printNGE(arr, N);
}
}
 
// This code is contributed by Code_Mech


Javascript




<script>
 
// JavaScript program for the above approach
 
// Function to find the NGE
function printNGE(A, n)
{
 
    // Formation of circular array
    let arr = Array.from({length: 2 * n}, (_, i) => 0);
     
    // Append the given array element twice
    for(let i = 0; i < 2 * n; i++)
        arr[i] = A[i % n];
 
    let next;
 
    // Iterate for all the
    // elements of the array
    for(let i = 0; i < n; i++)
    {
 
        // Initialise NGE as -1
        next = -1;
             
        for(let j = i + 1; j < 2 * n; j++)
        {
                 
            // Checking for next
            // greater element
            if (arr[i] < arr[j])
            {
                next = arr[j];
                break;
            }
        }
             
        // Print the updated NGE
        document.write(next + ", ");
    }
}
 
 
// Driver Code
 
    // Given array arr[]
    let arr = [ 1, 2, 1 ];
 
    let N = arr.length;
 
    // Function call
    printNGE(arr, N);
 
</script>


Output

2, -1, 2, 









This approach takes of O(n2) time but takes extra space of order O(n)

A space-efficient solution is to deal with circular arrays using the same array. If a careful observation is run through the array, then after the n-th index, the next index always starts from 0 so using the mod operator, we can easily access the elements of the circular list, if we use (i)%n and run the loop from i-th index to n+i-th index, and apply mod we can do the traversal in a circular array within the given array without using any extra space.

Below is the implementation of the above approach: 

C++




// C++ program to demonstrate the use of circular
// array without using extra memory space
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the Next Greater Element(NGE)
void printNGE(int A[], int n)
{
    int k;
    for(int i = 0; i < n; i++)
    {
        // Initialise k as -1 which is printed
        // when no NGE found
        k = -1;
        for(int j = i + 1; j < n + i; j++)
        {
            if (A[i] < A[j % n])
            {
                cout <<" "<< A[j % n];
                k = 1;
                break;
            }
        }
        // Gets executed when no NGE found
        if (k == -1)
            cout << "-1 ";
    }
}
 
// Driver Code
int main()
{
    // Given array arr[]
    int arr[] = { 8, 6, 7 };
    int N = sizeof(arr) / sizeof(arr[0]);
    // Function call
    printNGE(arr, N);
    return 0;
}
 
// This code is contributed by Aditya Kumar (adityakumar129)


C




// C program to demonstrate the use of circular
// array without using extra memory space
 
#include <stdio.h>
 
// Function to find the Next Greater Element(NGE)
void printNGE(int A[], int n)
{
    int k;
    for (int i = 0; i < n; i++) {
        // Initialise k as -1 which is printed when no NGE
        // found
        k = -1; //
        for (int j = i + 1; j < n + i; j++) {
            if (A[i] < A[j % n]) {
                printf("%d ", A[j % n]);
                k = 1;
                break;
            }
        }
        if (k == -1) // Gets executed when no NGE found
            printf("-1 ");
    }
}
 
// Driver Code
int main()
{
    // Given array arr[]
    int arr[] = { 8, 6, 7 };
    int N = sizeof(arr) / sizeof(arr[0]);
    // Function call
    printNGE(arr, N);
    return 0;
}


Java




// Java program to demonstrate the use of circular array
// without using extra memory space
import java.io.*;
 
class GFG {
 
    // Function to find the Next
    // Greater Element(NGE)
    static void printNGE(int A[], int n)
    {
        int k;
        for (int i = 0; i < n; i++) {
 
            // Initialise k as -1 which is printed when no
            // NGE found
            k = -1;
            for (int j = i + 1; j < n + i; j++) {
                if (A[i] < A[j % n]) {
                    System.out.print(A[j % n] + " ");
                    k = 1;
                    break;
                }
            }
            // Gets executed when no NGE found
            if (k == -1)
                System.out.print("-1 ");
        }
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        // Given array arr[]
        int[] arr = { 8, 6, 7 };
        int N = arr.length;
        // Function call
        printNGE(arr, N);
    }
}
 
// This code is contributed by Aditya Kumar (adityakumar129)


Python3




# Python3 program to demonstrate the use of circular
# array without using extra memory space
 
# Function to find the Next Greater Element(NGE)
def printNGE(A, n) :
 
    for i in range(n) :
       
        # Initialise k as -1 which is printed when no NGE
        # found
        k = -1
        for j in range(i + 1, n + i) :
            if (A[i] < A[j % n]) :
                print(A[j % n], end = " ")
                k = 1
                break
 
        if (k == -1) : # Gets executed when no NGE found
            print("-1 ", end = "")
 
# Given array arr[]
arr = [ 8, 6, 7 ]
 
N = len(arr)
 
# Function call
printNGE(arr, N)
 
# This code is contributed by divyeshrabadia07


C#




// C# program to demonstrate the
// use of circular array without
// using extra memory space
using System;
class GFG {
     
    // Function to find the Next
    // Greater Element(NGE)
    static void printNGE(int[] A, int n)
    {
        int k;
        for(int i = 0; i < n; i++)
        {
              
            // Initialise k as -1 which is
            // printed when no NGE found
            k = -1;
            for(int j = i + 1; j < n + i; j++)
            {
                if (A[i] < A[j % n])
                {
                    Console.Write(A[j % n] + " ");
                    k = 1;
                    break;
                }
            }
              
            // Gets executed when no NGE found
            if (k == -1)
                Console.Write("-1 ");
        }
    }
   
  static void Main()
  {
     
    // Given array arr[]
    int[] arr = { 8, 6, 7 };
  
    int N = arr.Length;
  
    // Function call
    printNGE(arr, N);
  }
}
 
// This code is contributed by divyesh072019


Javascript




<script>
 
    // JavaScript program to demonstrate the
    // use of circular array without
    // using extra memory space
     
    // Function to find the Next
    // Greater Element(NGE)
    function printNGE(A, n)
    {
        let k;
        for(let i = 0; i < n; i++)
        {
 
            // Initialise k as -1 which is
            // printed when no NGE found
            k = -1;
            for(let j = i + 1; j < n + i; j++)
            {
                if (A[i] < A[j % n])
                {
                    document.write(A[j % n] + " ");
                    k = 1;
                    break;
                }
            }
 
            // Gets executed when no NGE found
            if (k == -1)
                document.write("-1 ");
        }
    }
     
    // Given array arr[]
    let arr = [ 8, 6, 7 ];
  
    let N = arr.length;
  
    // Function call
    printNGE(arr, N);
 
</script>


Output

-1  7 8









 Time Complexity: O(n2
Auxiliary Space: O(1)

Method 3rd: The method uses the same concept used in method 2 for circular Array but uses Stack to find out the next greater element in O(n) time complexity where n is the size of the array. For better understanding, you can see the next greater element.

C++




#include <bits/stdc++.h>
using namespace std;
 
// Function to find the Next Greater Element(NGE)
void printNGE(int a[], int n)
{
    stack<int> s;
    vector<int> ans(n);
    for (int i = 2 * n - 1; i >= 0; i--) {
        while (!s.empty() && a[i % n] >= s.top())
            s.pop();
        if (i < n) {
            if (!s.empty())
                ans[i] = s.top();
            else
                ans[i] = -1;
        }
        s.push(a[i % n]);
    }
    for (int i = 0; i < n; i++)
        cout << ans[i] << " ";
}
 
// Driver Code
int main()
{
    int arr[] = { 8, 6, 7 };
    int N = sizeof(arr) / sizeof(arr[0]);
    printNGE(arr, N);
    return 0;
}


Java




/*package whatever //do not write package name here */
import java.io.*;
import java.util.*;
class GFG {
   
  public static void printNGE(int[] arr)
  {
        Stack<Integer> stack = new Stack<>();
        int n = arr.length;
        int[] result = new int[n];
 
        for(int i = 2*n - 1; i >= 0; i--)
        {
           
            // Remove all the elements in Stack that are less than arr[i%n]
            while(!stack.isEmpty() && arr[i % n] >= stack.peek()){
                stack.pop();
            }
            if(i < n)
            {
                if(!stack.isEmpty())
                    result[i] = stack.peek();
                else
                    result[i] = -1; // When none of elements in Stack are greater than arr[i%n]
            }
            stack.push(arr[i % n]);
        }
        for(int i:result)
        {
          System.out.print(i + " ");
        }
    }
   
  // Driver code
  public static void main (String[] args) {
      int[] arr = {8, 6 , 7};
       
      printNGE(arr);
       
    }
}
 
// This code is contributed by vaibhavpatel1904.


Python3




# Function to find the Next Greater Element(NGE)
def printNGE(a, n):
    s = []
    ans = [0] * n
    for i in range(2 * n - 1, -1, -1):
        while s and a[i % n] >= s[-1]:
            s.pop()
        if i < n:
            if s:
                ans[i] = s[-1]
 
            else:
                ans[i] = -1
 
        s.append(a[i % n])
 
    for i in range(n):
        print(ans[i], end=" ")
 
 
# Driver Code
if __name__ == "__main__":
    # Given array arr[]
    arr = [8, 6, 7]
 
    N = len(arr)
 
    # Function call
    printNGE(arr, N)


C#




// C# code for the above approach
 
using System;
using System.Collections;
 
public class GFG {
 
  // Function to find the Next Greater Element(NGE)
  static void printNGE(int[] arr)
  {
    Stack stack = new Stack();
    int n = arr.Length;
    int[] result = new int[n];
 
    for (int i = 2 * n - 1; i >= 0; i--) {
 
      // Remove all the elements in Stack that are
      // less than arr[i%n]
      while (stack.Count != 0
             && arr[i % n] >= (int)stack.Peek()) {
        stack.Pop();
      }
 
      if (i < n) {
        if (stack.Count != 0) {
          result[i] = (int)stack.Peek();
        }
        else {
          // When none of elements in Stack are
          // greater than arr[i%n]
          result[i] = -1;
        }
      }
      stack.Push(arr[i % n]);
    }
 
    foreach(int i in result) { Console.Write(i + " "); }
  }
 
  static public void Main()
  {
 
    // Code
    int[] arr = { 8, 6, 7 };
 
    printNGE(arr);
  }
}
 
// This code is contributed by lokesh.


Javascript




<script>
 
// Function to find the Next Greater Element(NGE)
function printNGE(a, n){
    let s = []
    let ans = new Array(n).fill(0)
    for(let i=2 * n - 1;i>=0;i--){
        while(s.length>0 && a[i % n] >= s[s.length - 1])
            s.pop()
        if(i < n){
            if(s.length>0)
                ans[i] = s[s.length-1]
 
            else
                ans[i] = -1
        }
 
        s.push(a[i % n])
    }
 
    for(let i=0;i<n;i++){
        document.write(ans[i]," ")
    }
}
 
 
// Driver Code
 
// Given array arr[]
let arr = [8, 6, 7]
 
let N = arr.length
 
// Function call
printNGE(arr, N)
 
// code is contributed by shinjanpatra
 
</script>


Output

-1 7 8 









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

Method :- 4 
In method 3, next greater element is calculated by traversing the array from backward (end) but we can also do the same in forward (start) traversal.

C++




// C++ code to find the next greater element
// in circular array.
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the Next Greater Element(NGE)
void printNGE(int nums[], int n)
{
    // Stores the next greater element for index i.
    vector<int> ans(n, -1);
    stack<int> s;
    for (int i = 0; i < 2 * n; i++) {
        while (!s.empty() && nums[s.top()] < nums[i % n]) {
            ans[s.top()] = nums[i % n];
            s.pop();
        }
        if (i < n)
            s.push(i);
    }
    for (auto it : ans)
        cout << it << " ";
}
 
// Driver Code
int main()
{
    int arr[] = { 8, 6, 7 };
    int N = sizeof(arr) / sizeof(arr[0]);
    printNGE(arr, N);
    return 0;
}
 
// This code is contributed by Aditya Kumar (adityakumar129)


Java




/*package whatever //do not write package name here */
import java.io.*;
import java.util.*;
class GFG {
 
  public static void printNGE(int[] arr)
  {
    Stack<Integer> stack = new Stack<>();
    int n = arr.length;
    int[] result = new int[n];
    Arrays.fill(result,-1);
 
    for(int i = 0; i <2*n-1; i++)
    {
 
      // Remove all the elements in Stack that are less than arr[i%n]
      while(!stack.isEmpty() && arr[i % n] > arr[stack.peek()]){
        result[stack.peek()] = arr[i%n];
        stack.pop();
      }
      if(i < n)
        stack.push(i);
    }
    for(int i:result)
    {
      System.out.print(i + " ");
    }
  }
 
  // Driver code
  public static void main (String[] args) {
    int[] arr = {8, 6 , 7};
 
    printNGE(arr);
 
  }
}
 
// This code is contributed by isha307.


Python3




# Python code for the above approach
def printNGE(arr):
    stack = []
    n = len(arr)
    result = [-1] * n
 
    for i in range(2 * n - 1):
       
        # Remove all the elements in Stack that are less than arr[i%n]
        while stack and arr[i % n] > arr[stack[-1]]:
            result[stack[-1]] = arr[i % n]
            stack.pop()
        if i < n:
            stack.append(i)
 
    for i in result:
        print(i, end=' ')
 
arr = [8, 6, 7]
printNGE(arr)
 
# This code is contributed by lokesh.


C#




// C# code for the above approach
using System;
using System.Collections;
using System.Collections.Generic;
 
public class GFG {
 
  static void printNGE(int[] arr)
  {
    Stack stack = new Stack();
    int n = arr.Length;
    int[] result = new int[n];
    Array.Fill(result, -1);
 
    for (int i = 0; i < 2 * n - 1; i++) {
 
      // Remove all the elements in Stack that are
      // less than arr[i%n]
      while (stack.Count != 0
             && arr[i % n] > arr[(int)stack.Peek()]) {
        result[(int)stack.Peek()] = arr[i % n];
        stack.Pop();
      }
      if (i < n)
        stack.Push(i);
    }
    foreach(int i in result) { Console.Write(i + " "); }
  }
 
  static public void Main()
  {
 
    // Code
    int[] arr = { 8, 6, 7 };
    printNGE(arr);
  }
}
 
// This code is contributed by lokeshmvs21.


Javascript




<script>
 
// JavaScript code to find the next greater element
// in circular array.
 
 
// Function to find the Next Greater Element(NGE)
function printNGE(nums, n)
{
    // Stores the next greater element for index i.
    let ans = new Array(n).fill(-1);
    let s = [];
    for (let i = 0; i < 2 * n; i++) {
        while (s.length > 0 && nums[s[s.length - 1]] < nums[i % n]) {
            ans[s[s.length - 1]] = nums[i % n];
            s.pop();
        }
        if (i < n)
            s.push(i);
    }
    for (let it of ans)
        document.write(it , " ");
}
 
// Driver Code
let arr = [ 8, 6, 7 ];
let N = arr.length;
printNGE(arr, N);
 
// This code is contributed by shinjanpatra
 
</script>


Output

-1 7 8 









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

Method 5: (Using Deque)

 Intuition:

We can use a stack to find the next greater element of each element in the given array. We loop through the array twice to compare each element with every other element. For each element, we push its index onto the stack. If the current element is greater than the top element of the stack, we pop the stack and update the result array with the current element for the popped index. We continue this process until the stack is empty or the top element of the stack is greater than or equal to the current element. Finally, we return the result array.

C++




// C++ code to find the next greater element
// in circular array.
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the Next Greater Element(NGE)
void printNGE(vector<int>& nums,vector<int> &ret) 
    {
  //declare a deque for accessing the elements
        deque<int> dq;
        //traverse throgh the entire vector and push the element into the deque
        for(int i = 1; i < nums.size(); i++)
            dq.push_back(nums[i]);
        //check for the existence using two loops
        for(int i = 0; i < nums.size(); i++)
        {
            int n = ret.size();
            for(int j = 0; j < dq.size(); j++)
            {
                if(dq[j] > nums[i])
                {
                    ret.emplace_back(dq[j]);
                    break;
                }
            }
            if(ret.size() == n)
                ret.emplace_back(-1); //push -1 at the end of vector
            dq.pop_front(); //pop if doesn'match
            dq.push_back(nums[i]); //push if matches
        }
    }
// Driver Code
int main()
{
    vector<int> arr{ 8, 6, 7 };
  //store the result in the res vector
    vector<int> res;
    printNGE(arr,res);
      for(int i=0;i<res.size();i++){
        cout<<res[i]<<" ";
    }
    return 0;
}


Java




import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Deque;
import java.util.List;
 
class GFG {
    // Function to find the Next Greater Element (NGE) for each
   // element in the list 'nums'
    public static void printNGE(List<Integer> nums, List<Integer> ret) {
        // Create a deque (double-ended queue) to store elements for
      // finding NGE
        Deque<Integer> dq = new ArrayDeque<>();
         
        // Add all elements from index 1 to the deque (except for the first element)
        // We start from index 1 because there is no element to the left of the first element.
        for (int i = 1; i < nums.size(); i++) {
            dq.add(nums.get(i));
        }
         
        // Iterate through each element in 'nums' to find its NGE
        for (int i = 0; i < nums.size(); i++) {
            int n = ret.size();
            // Iterate through the elements in the deque to
          // find the NGE for the current element in 'nums'
            for (int j = 0; j < dq.size(); j++) {
                // If the top element of the deque is greater than
              // the current element in 'nums',
                // it is the Next Greater Element, so add it to the
              // 'ret' list and break the loop.
                if (dq.peek() > nums.get(i)) {
                    ret.add(dq.peek());
                    break;
                }
                // If the top element of the deque is not the NGE,
              // move it to the end of the deque
                dq.add(dq.poll());
            }
             
            // If the NGE is not found (the loop didn't break),
          // add -1 to the 'ret' list as there is no greater element.
            if (ret.size() == n) {
                ret.add(-1);
            }
             
            // Remove the first element from the deque
          // (which was used for finding the NGE)
            dq.pollFirst();
             
            // Add the current element from 'nums' to the end
          // of the deque for the next iteration.
            dq.add(nums.get(i));
        }
    }
 
    public static void main(String[] args) {
        List<Integer> arr = List.of(8, 6, 7);
        List<Integer> res = new ArrayList<>();
         
        // Find and store the Next Greater Element (NGE) for each
      // element in 'arr' in the 'res' list
        printNGE(arr, res);
         
        // Print the 'res' list containing the Next Greater
      // Element (NGE) for each element in 'arr'
        for (int i = 0; i < res.size(); i++) {
            System.out.print(res.get(i) + " ");
        }
    }
}


Python3




# Python code to find the next greater element
# in circular array using deque
 
# importing the module
from collections import deque
 
 
# main function which is finding the
# next greater element (NGE)
def print_NGE(nums):
    ret = []
     
    # initializing a deque
    dq = deque()
     
    # iterating through the list
    # and adding elements in deque
    for i in range(1, len(nums)):
        dq.append(nums[i])
     
    # checking the existence of
    # duplicate values
    for i in range(len(nums)):
        n = len(ret)
        for j in range(len(dq)):
            if dq[j] > nums[i]:
                ret.append(dq[j])
                break
         
        # appending -1 at the end
        if len(ret) == n:
            ret.append(-1)
         
        # popping if it doesn't match
        dq.popleft()
         
        # pushing if it matches
        dq.append(nums[i])
         
    #returning the final result
    return ret
 
# Driver Code
arr = [8, 6, 7]
res = print_NGE(arr)
for i in range(len(res)):
    print(res[i], end=" ")


C#




using System;
using System.Collections.Generic;
 
public class GFG
{
    // Function to find the Next Greater Element (NGE)
    static void PrintNGE(List<int> nums, List<int> ret)
    {
        // Declare a deque for accessing the elements
        Queue<int> dq = new Queue<int>(nums.GetRange(1, nums.Count - 1));
         
        // Traverse through the entire list and find the NGE
        for (int i = 0; i < nums.Count; i++)
        {
            int n = ret.Count;
            foreach (int num in dq)
            {
                if (num > nums[i])
                {
                    ret.Add(num);
                    break;
                }
            }
             
            if (ret.Count == n)
                ret.Add(-1); // Push -1 at the end of the list
 
            dq.Dequeue(); // Pop if it doesn't match
            dq.Enqueue(nums[i]); // Push if it matches
        }
    }
 
    // Driver Code
    public static void Main(string[] args)
    {
        List<int> arr = new List<int> { 8, 6, 7 };
        // Store the result in the 'res' list
        List<int> res = new List<int>();
        PrintNGE(arr, res);
 
        foreach (int val in res)
        {
            Console.Write(val + " ");
        }
    }
}


Javascript




// Function to find the Next Greater Element(NGE)
function printNGE(nums, ret) {
  // Declare a deque for accessing the elements
  let dq = [];
   
  // Traverse through the entire array and push the elements into the deque
  for(let i = 1; i < nums.length; i++) {
    dq.push(nums[i]);
  }
   
  // Check for the existence using two loops
  for(let i = 0; i < nums.length; i++) {
    let n = ret.length;
     
    for(let j = 0; j < dq.length; j++) {
      if(dq[j] > nums[i]) {
        ret.push(dq[j]);
        break;
      }
    }
     
    if(ret.length === n) {
      ret.push(-1); // Push -1 at the end of array
    }
     
    dq.shift(); // Pop if doesn't match
    dq.push(nums[i]); // Push if matches
  }
}
 
// Driver Code
let arr = [8, 6, 7];
let res = [];
 
printNGE(arr, res);
 
for(let i = 0; i < res.length; i++) {
  console.log(res[i] + " ");
}


Output

-1 7 8 









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



Last Updated : 02 Nov, 2023
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