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

Last Updated : 28 Aug, 2023

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[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:

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

Steps to implement-

• Declare a variable count and initialize it with 0
• Traverse the array to pick elements one by one
• After picking an element traverse to its left and if k right element exists then traverse k element to its right
• If our element is greater than all element in its left and k elements in its right then increment the count by 1
• In last value stored in count will be our answer

Code-

## C++

 `// C++ Program to implement` `// the above approach` `#include ` `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)` `{` `  ``//To store final answer` `  ``int` `count=0;` `  `  `  ``//pick element one by one` `  ``for``(``int` `i=0;i=0){` `          ``//when our element is not greater than an element on its left then break the inner loop` `          ``if``(arr[i]<=arr[j]){``break``;}` `          ``j--;` `      ``}` `      `  `      ``//Traverse towards its right` `      ``while``(i+m

## Java

 `// Java program for the above approach` `import` `java.util.*;`   `public` `class` `Main {`   `    ``// Function to print the count of` `    ``// Array elements greater than all` `    ``// elements on its left and next K` `    ``// elements on its right` `    ``public` `static` `int` `countElements(``int` `arr[], ``int` `n, ``int` `k)` `    ``{` `        ``// To store final answer` `        ``int` `count = ``0``;`   `        ``// pick element one by one` `        ``for` `(``int` `i = ``0``; i < n; i++) {` `            ``// For traversal in left` `            ``int` `j = i - ``1``;` `            ``// For traversal in right` `            ``int` `m = ``1``;`   `            ``// Traverse towards its left` `            ``while` `(j >= ``0``) {` `                ``// when our element is not` `                ``// greater than an element on` `                ``// its left then break the inner loop` `                ``if` `(arr[i] <= arr[j]) {` `                    ``break``;` `                ``}` `                ``j--;` `            ``}`   `            ``// Traverse towards its right` `            ``while` `(i + m < n && m <= k) {`   `                ``// when our element not greater` `                ``// than an element in its k right` `                ``// element then break the inner loop` `                ``if` `(arr[i] <= arr[i + m]) {` `                    ``break``;` `                ``}` `                ``m++;` `            ``}`   `            ``// When our element is greater than` `            ``// all elements on it left and` `            ``// is greater than k element on its right` `            ``if` `(j == -``1``) {` `                ``if` `(i + m == n) {` `                    ``count++;` `                ``}` `                ``else` `if` `(m == k + ``1``) {` `                    ``count++;` `                ``}` `            ``}` `        ``}` `        ``return` `count;` `    ``}`   `    ``// Driver Code` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``int` `arr[] = { ``3``, ``1``, ``2``, ``7``, ``5``, ``1``, ``2``, ``6` `};` `        ``int` `K = ``2``;` `        ``int` `n = arr.length;` `        ``System.out.println(countElements(arr, n, K));` `    ``}` `}`

## Python3

 `# 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):`   `    ``# To store final answer` `    ``count ``=` `0`   `    ``# pick element one by one` `    ``for` `i ``in` `range``(n):`   `        ``# For traversal in left` `        ``j ``=` `i``-``1`   `        ``# For traversal in right` `        ``m ``=` `1`   `        ``# Traverse towards its left` `        ``while` `j >``=` `0``:` `            ``# when our element is not greater than an element on its left then break the inner loop` `            ``if` `arr[i] <``=` `arr[j]:` `                ``break` `            ``j ``-``=` `1`   `        ``# Traverse towards its right` `        ``while` `i``+``m < n ``and` `m <``=` `k:` `            ``# when our element not greater than an element in its k right element then break the inner loop` `            ``if` `arr[i] <``=` `arr[i``+``m]:` `                ``break` `            ``m ``+``=` `1`   `        ``# When our element is greater than all elements on it left and` `        ``# is greater than k element on its right` `        ``if` `j ``=``=` `-``1``:` `            ``if` `i``+``m ``=``=` `n:` `                ``count ``+``=` `1` `            ``elif` `m ``=``=` `k``+``1``:` `                ``count ``+``=` `1`   `    ``return` `count`   `# Driver Code` `if` `__name__ ``=``=` `'__main__'``:` `    ``arr ``=` `[``3``, ``1``, ``2``, ``7``, ``5``, ``1``, ``2``, ``6``]` `    ``K ``=` `2` `    ``n ``=` `len``(arr)` `    ``print``(countElements(arr, n, K))`

## C#

 `using` `System;`   `class` `Program` `{` `    ``// 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)` `    ``{` `        ``//To store final answer` `        ``int` `count = 0;`   `        ``//pick element one by one` `        ``for` `(``int` `i = 0; i < n; i++)` `        ``{` `            ``//For traversal in left` `            ``int` `j = i - 1;` `            ``//For traversal in right` `            ``int` `m = 1;`   `            ``//Traverse towards its left` `            ``while` `(j >= 0)` `            ``{` `                ``//when our element is not greater than an element on its left then break the inner loop` `                ``if` `(arr[i] <= arr[j]) { ``break``; }` `                ``j--;` `            ``}`   `            ``//Traverse towards its right` `            ``while` `(i + m < n && m <= k)` `            ``{` `                ``//when our element not greater than an element in its k right element then break the inner loop` `                ``if` `(arr[i] <= arr[i + m]) { ``break``; }` `                ``m++;` `            ``}`   `            ``//When our element is greater than all elements on it left and ` `            ``//is greater than k element on its right` `            ``if` `(j == -1)` `            ``{` `                ``if` `(i + m == n) { count++; }` `                ``else` `if` `(m == k + 1) { count++; }` `            ``}`   `        ``}` `        ``return` `count;` `    ``}`   `    ``static` `void` `Main(``string``[] args)` `    ``{` `        ``int``[] arr = { 3, 1, 2, 7, 5, 1, 2, 6 };` `        ``int` `K = 2;` `        ``int` `n = arr.Length;` `        ``Console.WriteLine(countElements(arr, n, K));` `    ``}` `}` `// This code is contributed by rudra1807raj`

## Javascript

 `// 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, k) {` `    ``// To store final answer` `    ``let count = 0;`   `    ``// Pick element one by one` `    ``for` `(let i = 0; i < arr.length; i++) {` `        ``// For traversal in left` `        ``let j = i - 1;` `        ``// For traversal in right` `        ``let m = 1;`   `        ``// Traverse towards its left` `        ``while` `(j >= 0) {` `            ``// When our element is not greater than an element on its left then break the inner loop` `            ``if` `(arr[i] <= arr[j]) {` `                ``break``;` `            ``}` `            ``j--;` `        ``}`   `        ``// Traverse towards its right` `        ``while` `(i + m < arr.length && m <= k) {` `            ``// When our element not greater than an element in its k right element then break the inner loop` `            ``if` `(arr[i] <= arr[i + m]) {` `                ``break``;` `            ``}` `            ``m++;` `        ``}`   `        ``// When our element is greater than all elements on its left and ` `        ``// is greater than k elements on its right` `        ``if` `(j === -1) {` `            ``if` `(i + m === arr.length) {` `                ``count++;` `            ``} ``else` `if` `(m === k + 1) {` `                ``count++;` `            ``}` `        ``}` `    ``}` `    ``return` `count;` `}`   `// Driver Code` `const arr = [3, 1, 2, 7, 5, 1, 2, 6];` `const K = 2;` `console.log(countElements(arr, K));`

Output

```2

```

Time Complexity: O(N2), because of one external loop and two inner loop means nested loop
Auxiliary Space: O(1), because no extra space has been used
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` `// 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 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` `# 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

 ``

Output

```2

```

Time Complexity: O(N), The given program uses a stack to find the next greater element for each element in the array. The time complexity of this operation is O(n) because each element is pushed and popped from the stack only once. After finding the next greater element for each element, the program iterates over the array to count the elements that satisfy the given condition. This operation also takes O(n) time. Therefore, the overall time complexity of the program is O(n).
Auxiliary Space: O(N), The program uses two additional data structures: a stack and a vector. The size of the stack can be at most n because each element can be pushed and popped from the stack only once. The size of the vector is also n because it stores the index of the next greater element for each element. Therefore, the total space complexity of the program is O(n).

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