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Count subarrays with all elements greater than K

  • Difficulty Level : Easy
  • Last Updated : 07 May, 2021

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
?>

Javascript




<script>
// javascript program to prvar the number of subarrays such
// that all elements are greater than K    
// Function to count number of subarrays
    function countSubarrays(a , n , x) {
        var count = 0;
 
        var 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 != 0)
            number += (count) * (count + 1) / 2;
 
        return number;
    }
 
    // Driver Code
     
        var a = [ 3, 4, 5, 6, 7, 2, 10, 11 ];
        var n = a.length;
        var k = 5;
 
        document.write(countSubarrays(a, n, k));
 
// This code is contributed by todaysgaurav
</script>
Output: 
6

 

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




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