Count of Array elements greater than all elements on its left and next K elements on its right

Given an array arr[], the task is to print the number of elements which are greater than all the elements on its left as well as greater than the next K elements on its right.

Examples:

Input: arr[] = { 4, 2, 3, 6, 4, 3, 2}, K = 2
Output: 2
Explanation:
arr[0](= 4): arr[0] is the 1st element in the array and greater than its next K(= 2) elements {2, 3}.
arr[2](= 6): arr[2] is greater than all elements on its left {4, 2, 3} and greater than its next K(= 2) elements {4, 3}.
Therefore, only two elements satisfy the given condition.

Input: arr[] = { 3, 1, 2, 7, 5, 1, 2, 6}, K = 2
Output: 2

Naive Approach:
Traverse over the array and for each element, check if all elements on its left are smaller than it as well as the next K elements on its right are smaller than it. For every such element, increase count. Finally, print count.

Time Complexity: O(N²)
Auxiliary Space: O(1)

Efficient Approach:
The above approach can be optimized by using the Stack Data Structure. Follow the steps below to solve the problem:

Initialize a new array and store the index of the Next Greater Element for each array element using Stack.
Traverse the given array and for each element, check if it is maximum obtained so far and its next greater element is at least K indices after the current index. If found to be true, increase count.
Finally, print count.

Below is the implementation of the above approach:

C++

// C++ Program to implement 
// the above approach 
#include <bits/stdc++.h> 
using namespace std; 
  
// Function to print the count of 
// Array elements greater than all 
// elemnts on its left and next K 
// elements on its right 
int countElements(int arr[], int n, 
                  int k) 
{ 
  
    stack<int> s; 
  
    vector<int> next_greater(n, n + 1); 
  
    // Iterate over the array 
    for (int i = 0; i < n; i++) { 
  
        if (s.empty()) { 
            s.push(i); 
            continue; 
        } 
  
        // If the stack is not empty and 
        // the element at the top of the 
        // stack is smaller than arr[i] 
        while (!s.empty() 
               && arr[s.top()] < arr[i]) { 
            // Store the index of next 
            // greater element 
            next_greater[s.top()] = i; 
  
            // Pop the top element 
            s.pop(); 
        } 
  
        // Insert the current index 
        s.push(i); 
    } 
  
    // Stores the count 
    int count = 0; 
    int maxi = INT_MIN; 
    for (int i = 0; i < n; i++) { 
        if (next_greater[i] - i > k 
            && maxi < arr[i]) { 
            maxi = max(maxi, arr[i]); 
            count++; 
        } 
    } 
  
    return count; 
} 
  
// Driver Code 
int main() 
{ 
  
    int arr[] = { 4, 2, 3, 6, 4, 3, 2 }; 
    int K = 2; 
    int n = sizeof(arr) / sizeof(arr[0]); 
    cout << countElements(arr, n, K); 
  
    return 0; 
} 

Java

// Java program to implement 
// the above approach 
import java.util.*; 
  
class GFG{ 
  
// Function to print the count of 
// Array elements greater than all 
// elemnts on its left and next K 
// elements on its right 
static int countElements(int arr[], int n, 
                                    int k) 
{ 
    Stack<Integer> s = new Stack<Integer>(); 
  
    int []next_greater = new int[n + 1]; 
    Arrays.fill(next_greater, n); 
  
    // Iterate over the array 
    for(int i = 0; i < n; i++) 
    { 
        if (s.isEmpty()) 
        { 
            s.add(i); 
            continue; 
        } 
  
        // If the stack is not empty and 
        // the element at the top of the 
        // stack is smaller than arr[i] 
        while (!s.isEmpty() &&  
               arr[s.peek()] < arr[i])  
        { 
              
            // Store the index of next 
            // greater element 
            next_greater[s.peek()] = i; 
  
            // Pop the top element 
            s.pop(); 
        } 
  
        // Insert the current index 
        s.add(i); 
    } 
  
    // Stores the count 
    int count = 0; 
    int maxi = Integer.MIN_VALUE; 
      
    for(int i = 0; i < n; i++)  
    { 
        if (next_greater[i] - i > k &&  
              maxi < arr[i]) 
        { 
            maxi = Math.max(maxi, arr[i]); 
            count++; 
        } 
    } 
    return count; 
} 
  
// Driver Code 
public static void main(String[] args) 
{ 
    int arr[] = { 4, 2, 3, 6, 4, 3, 2 }; 
    int K = 2; 
    int n = arr.length; 
      
    System.out.print(countElements(arr, n, K)); 
} 
} 
  
// This code is contributed by PrinciRaj1992 

Python3

   
# Python3 program to implement 
# the above approach 
import sys 
  
# Function to print the count of 
# Array elements greater than all 
# elemnts on its left and next K 
# elements on its right 
def countElements(arr, n, k): 
  
    s = [] 
  
    next_greater = [n] * (n + 1) 
  
    # Iterate over the array 
    for i in range(n): 
        if(len(s) == 0): 
            s.append(i) 
            continue
  
        # If the stack is not empty and 
        # the element at the top of the 
        # stack is smaller than arr[i] 
        while(len(s) != 0 and
              arr[s[-1]] < arr[i]): 
                    
            # Store the index of next 
            # greater element 
            next_greater[s[-1]] = i 
  
            # Pop the top element 
            s.pop(-1) 
  
        # Insert the current index 
        s.append(i) 
  
    # Stores the count  
    count = 0
    maxi = -sys.maxsize - 1
      
    for i in range(n): 
        if(next_greater[i] - i > k and
             maxi < arr[i]): 
            maxi = max(maxi, arr[i]) 
            count += 1
  
    return count 
  
# Driver Code 
if __name__ == '__main__': 
  
    arr = [ 4, 2, 3, 6, 4, 3, 2 ] 
    K = 2
    n = len(arr) 
  
    # Function call 
    print(countElements(arr, n, K)) 
  
# This code is contributed by Shivam Singh 

C#

// C# program to implement 
// the above approach 
using System; 
using System.Collections.Generic; 
  
class GFG{ 
      
// Function to print the count of  
// Array elements greater than all  
// elemnts on its left and next K  
// elements on its right  
static int countElements(int[] arr, int n, 
                         int k)  
{  
    Stack<int> s = new Stack<int>();  
  
    int[] next_greater = new int[n + 1];  
    Array.Fill(next_greater, n);  
  
    // Iterate over the array  
    for(int i = 0; i < n; i++)  
    {  
        if (s.Count == 0)  
        {  
            s.Push(i);  
            continue;  
        }  
  
        // If the stack is not empty and  
        // the element at the top of the  
        // stack is smaller than arr[i]  
        while (s.Count != 0 &&  
               arr[s.Peek()] < arr[i])  
        {  
              
            // Store the index of next  
            // greater element  
            next_greater[s.Peek()] = i;  
  
            // Pop the top element  
            s.Pop();  
        }  
  
        // Insert the current index  
        s.Push(i);  
    }  
  
    // Stores the count  
    int count = 0;  
    int maxi = Int32.MinValue;  
      
    for(int i = 0; i < n; i++)  
    {  
        if (next_greater[i] - i > k &&  
                       maxi < arr[i])  
        {  
            maxi = Math.Max(maxi, arr[i]);  
            count++;  
        }  
    }  
    return count;  
}  
  
// Driver Code 
static void Main() 
{ 
    int[] arr = { 4, 2, 3, 6, 4, 3, 2 };  
    int K = 2;  
    int n = arr.Length;  
  
    Console.Write(countElements(arr, n, K)); 
} 
}  
  
// This code is contributed by divyeshrabadiya07 

Output:

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

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My Personal Notes

C++

Java

Python3

C#

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