# 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:
Explanation:
arr(= 4): arr is the 1st element in the array and greater than its next K(= 2) elements {2, 3}.
arr(= 6): arr 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:

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 ` `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); ` `    ``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 s = ``new` `Stack(); ` ` `  `    ``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:

```2
```

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

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