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: 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.
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 <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)
{
//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
int main()
{
int arr[] = { 3, 1, 2, 7, 5, 1, 2, 6 };
int K = 2;
int n = sizeof(arr) / sizeof(arr[0]);
cout << countElements(arr, n, K);
return 0;
}
// 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));
}
}
# 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))
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
// 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));
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:
- 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
// 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 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 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# 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
<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>
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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