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: 

1. Initialize a new array and store the index of the Next Greater Element for each array element using Stack.
2. 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.
3. 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
// elements 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
// elements 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
# elements 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
// elements 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

Javascript

<script>

      // JavaScript program to implement
      // the above approach

      // Function to print the count of
      // Array elements greater than all
      // elements on its left and next K
      // elements on its right
      function countElements(arr, n, k) {
        var s = [];

        var next_greater = new Array(n + 1).fill(n);

        // Iterate over the array
        for (var i = 0; i < n; i++) {
          if (s.length === 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.length !== 0 && arr[s[s.length - 1]] <
          arr[i]) {
            // Store the index of next
            // greater element
            next_greater[s[s.length - 1]] = i;

            // Pop the top element
            s.pop();
          }

          // Insert the current index
          s.push(i);
        }

        // Stores the count
        var count = 0;
        //min integer value
        var maxi = -2147483648;

        for (var 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
      var arr = [4, 2, 3, 6, 4, 3, 2];
      var K = 2;
      var n = arr.length;

      document.write(countElements(arr, n, K));

</script>

Output:

2

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