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.
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++
#include <bits/stdc++.h>
using namespace std;
int countElements( int arr[], int n,
int k)
{
int count=0;
for ( int i=0;i<n;i++){
int j=i-1;
int m=1;
while (j>=0){
if (arr[i]<=arr[j]){ break ;}
j--;
}
while (i+m<n && m<=k){
if (arr[i]<=arr[i+m]){ break ;}
m++;
}
if (j==-1){
if (i+m==n){count++;}
else if (m==k+1){count++;}
}
}
return count;
}
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
import java.util.*;
public class Main {
public static int countElements( int arr[], int n, int k)
{
int count = 0 ;
for ( int i = 0 ; i < n; i++) {
int j = i - 1 ;
int m = 1 ;
while (j >= 0 ) {
if (arr[i] <= arr[j]) {
break ;
}
j--;
}
while (i + m < n && m <= k) {
if (arr[i] <= arr[i + m]) {
break ;
}
m++;
}
if (j == - 1 ) {
if (i + m == n) {
count++;
}
else if (m == k + 1 ) {
count++;
}
}
}
return count;
}
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
def countElements(arr, n, k):
count = 0
for i in range (n):
j = i - 1
m = 1
while j > = 0 :
if arr[i] < = arr[j]:
break
j - = 1
while i + m < n and m < = k:
if arr[i] < = arr[i + m]:
break
m + = 1
if j = = - 1 :
if i + m = = n:
count + = 1
elif m = = k + 1 :
count + = 1
return count
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
{
static int countElements( int [] arr, int n, int k)
{
int count = 0;
for ( int i = 0; i < n; i++)
{
int j = i - 1;
int m = 1;
while (j >= 0)
{
if (arr[i] <= arr[j]) { break ; }
j--;
}
while (i + m < n && m <= k)
{
if (arr[i] <= arr[i + m]) { break ; }
m++;
}
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));
}
}
|
Javascript
function countElements(arr, k) {
let count = 0;
for (let i = 0; i < arr.length; i++) {
let j = i - 1;
let m = 1;
while (j >= 0) {
if (arr[i] <= arr[j]) {
break ;
}
j--;
}
while (i + m < arr.length && m <= k) {
if (arr[i] <= arr[i + m]) {
break ;
}
m++;
}
if (j === -1) {
if (i + m === arr.length) {
count++;
} else if (m === k + 1) {
count++;
}
}
}
return count;
}
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++
#include <bits/stdc++.h>
using namespace std;
int countElements( int arr[], int n,
int k)
{
stack< int > s;
vector< int > next_greater(n, n + 1);
for ( int i = 0; i < n; i++) {
if (s.empty()) {
s.push(i);
continue ;
}
while (!s.empty()
&& arr[s.top()] < arr[i]) {
next_greater[s.top()] = i;
s.pop();
}
s.push(i);
}
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;
}
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
import java.util.*;
class GFG{
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);
for ( int i = 0 ; i < n; i++)
{
if (s.isEmpty())
{
s.add(i);
continue ;
}
while (!s.isEmpty() &&
arr[s.peek()] < arr[i])
{
next_greater[s.peek()] = i;
s.pop();
}
s.add(i);
}
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;
}
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));
}
}
|
Python3
import sys
def countElements(arr, n, k):
s = []
next_greater = [n] * (n + 1 )
for i in range (n):
if ( len (s) = = 0 ):
s.append(i)
continue
while ( len (s) ! = 0 and
arr[s[ - 1 ]] < arr[i]):
next_greater[s[ - 1 ]] = i
s.pop( - 1 )
s.append(i)
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
if __name__ = = '__main__' :
arr = [ 4 , 2 , 3 , 6 , 4 , 3 , 2 ]
K = 2
n = len (arr)
print (countElements(arr, n, K))
|
C#
using System;
using System.Collections.Generic;
class GFG{
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);
for ( int i = 0; i < n; i++)
{
if (s.Count == 0)
{
s.Push(i);
continue ;
}
while (s.Count != 0 &&
arr[s.Peek()] < arr[i])
{
next_greater[s.Peek()] = i;
s.Pop();
}
s.Push(i);
}
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;
}
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));
}
}
|
Javascript
<script>
function countElements(arr, n, k) {
var s = [];
var next_greater = new Array(n + 1).fill(n);
for ( var i = 0; i < n; i++) {
if (s.length === 0) {
s.push(i);
continue ;
}
while (s.length !== 0 && arr[s[s.length - 1]] <
arr[i]) {
next_greater[s[s.length - 1]] = i;
s.pop();
}
s.push(i);
}
var count = 0;
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;
}
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).
Last Updated :
28 Aug, 2023
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