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Find the next greater element in a Circular Array

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  • Difficulty Level : Medium
  • Last Updated : 29 Apr, 2022

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)

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)

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)


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