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Count subarrays with all elements greater than K
  • Difficulty Level : Easy
  • Last Updated : 09 Jan, 2019
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Given an array of N integers and a number K, the task is to find the number of subarrays such that all elements are greater than K in it.

Examples:

Input: a[] = {3, 4, 5, 6, 7, 2, 10, 11}, K = 5
Output: 6
The possible subarrays are {6}, {7}, {6, 7}, {10}, {11} and {10, 11}.

Input: a[] = {8, 25, 10, 19, 19, 18, 20, 11, 18}, K = 13
Output: 12

Approach: The idea is to start traversing the array using a counter. If the current element is greater than the given value X, increment the counter otherwise add counter*(counter+1)/2 to the number of subarrays and reinitialize counter to 0.



Below is the implementation of the above approach:

C++




// C++ program to print the number of subarrays such
// that all elements are greater than K
#include <bits/stdc++.h>
using namespace std;
  
// Function to count number of subarrays
int countSubarrays(int a[], int n, int x)
{
    int count = 0;
  
    int number = 0;
  
    // Iterate in the array
    for (int i = 0; i < n; i++) {
  
        // check if array element
        // greater then X or not
        if (a[i] > x) {
            count += 1;
        }
        else {
  
            number += (count) * (count + 1) / 2;
            count = 0;
        }
    }
  
    // After iteration complete
    // check for the last set of subarrays
    if (count)
        number += (count) * (count + 1) / 2;
  
    return number;
}
  
// Driver Code
int main()
{
    int a[] = { 3, 4, 5, 6, 7, 2, 10, 11 };
    int n = sizeof(a) / sizeof(a[0]);
    int k = 5;
  
    cout << countSubarrays(a, n, k);
  
    return 0;
}

Java




// Java program to print the number of subarrays such 
// that all elements are greater than K 
  
import java.io.*;
  
class GFG {
      
// Function to count number of subarrays 
static int countSubarrays(int a[], int n, int x) 
    int count = 0
  
    int number = 0
  
    // Iterate in the array 
    for (int i = 0; i < n; i++) { 
  
        // check if array element 
        // greater then X or not 
        if (a[i] > x) { 
            count += 1
        
        else
  
            number += (count) * (count + 1) / 2
            count = 0
        
    
  
    // After iteration complete 
    // check for the last set of subarrays 
    if (count!=0
        number += (count) * (count + 1) / 2
  
    return number; 
  
// Driver Code 
    public static void main (String[] args) {
        int a[] = { 3, 4, 5, 6, 7, 2, 10, 11 }; 
        int n = a.length; 
        int k = 5
  
        System.out.println (countSubarrays(a, n, k)); 
          
    }
}

Python3




# Python program to print the number of 
# subarrays such that all elements are 
# greater than K
  
# Function to count number of subarrays
def countSubarrays(a, n, x):
    count = 0
    number = 0
      
    # Iterate in the array
    for i in range(n):
          
        # check if array element greater
        # then X or not
        if (a[i] > x):
            count += 1
        else:
            number += (count) * (count + 1) / 2
            count = 0
              
    # After iteration complete check for
    # the last set of subarrays
    if (count):
        number += (count) * (count + 1) / 2
    return int(number)
  
# Driver Code
if __name__ == '__main__':
    a = [3, 4, 5, 6, 7, 2, 10, 11]
    n = len(a)
    k = 5
    print(countSubarrays(a, n, k))
  
# This code is contributed by 29AjayKumar

C#




// C# program to print the number of subarrays such 
// that all elements are greater than K 
  
using System;
  
class GFG {
      
// Function to count number of subarrays 
static int countSubarrays(int []a, int n, int x) 
    int count = 0; 
  
    int number = 0; 
  
    // Iterate in the array 
    for (int i = 0; i < n; i++) { 
  
        // check if array element 
        // greater then X or not 
        if (a[i] > x) { 
            count += 1; 
        
        else
  
            number += (count) * (count + 1) / 2; 
            count = 0; 
        
    
  
    // After iteration complete 
    // check for the last set of subarrays 
    if (count!=0) 
        number += (count) * (count + 1) / 2; 
  
    return number; 
  
// Driver Code 
    public static void Main () {
        int []a = { 3, 4, 5, 6, 7, 2, 10, 11 }; 
        int n = a.Length; 
        int k = 5; 
  
        Console.WriteLine(countSubarrays(a, n, k)); 
          
    }
}
// This code is contributed by anuj_67..

PHP




<?php
// PHP program to print the number 
// of subarrays such that all 
// elements are greater than K
  
// Function to count number 
// of subarrays
function countSubarrays($a, $n, $x)
{
    $count = 0; $number = 0;
  
    // Iterate in the array
    for ($i = 0; $i < $n; $i++) 
    {
  
        // check if array element
        // greater then X or not
        if ($a[$i] > $x)
        {
            $count += 1;
        }
        else 
        {
            $number += ($count) * 
                       ($count + 1) / 2;
            $count = 0;
        }
    }
  
    // After iteration complete
    // check for the last set
    // of subarrays
    if ($count)
        $number += ($count) *
                   ($count + 1) / 2;
  
    return $number;
}
  
// Driver Code
$a = array(3, 4, 5, 6, 7, 2, 10, 11);
$n = count($a);
$k = 5;
  
echo countSubarrays($a, $n, $k);
  
// This code is contributed by anuj_67
?>
Output:
6

Time Complexity: O(N)
Auxiliary Space: O(1)

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