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# Count of subarrays whose maximum element is greater than k

• Difficulty Level : Medium
• Last Updated : 20 May, 2021

Given an array of n elements and an integer k. The task is to find the count of subarray which has maximum element greater than K.
Examples :

Input : arr[] = {1, 2, 3} and k = 2.
Output : 3
All the possible subarrays of arr[] are
{ 1 }, { 2 }, { 3 }, { 1, 2 }, { 2, 3 },
{ 1, 2, 3 }.
Their maximum elements are 1, 2, 3, 2, 3, 3.
There are only 3 maximum elements > 2.

The idea is to approach problem by counting subarrays whose maximum element is less than or equal to k as counting such subarrays is easier. To find the number of subarray whose maximum element is less than or equal to k, remove all the element which is greater than K and find the number of subarray with the left elements.
Once we find above count, we can subtract it from n*(n+1)/2 to get our required result. Observe, there can be n*(n+1)/2 possible number of subarray of any array of size n. So, finding the number of subarray whose maximum element is less than or equal to K and substracting it from n*(n+1)/2 gets us the answer.
Below is the implementation of this approach:

## C++

 // C++ program to count number of subarrays// whose maximum element is greater than K.#include using namespace std; // Return number of subarrays whose maximum// element is less than or equal to K.int countSubarray(int arr[], int n, int k){    // To store count of subarrays with all    // elements less than or equal to k.    int s = 0;     // Traversing the array.    int i = 0;    while (i < n) {        // If element is greater than k, ignore.        if (arr[i] > k) {            i++;            continue;        }         // Counting the subarray length whose        // each element is less than equal to k.        int count = 0;        while (i < n && arr[i] <= k) {            i++;            count++;        }         // Suming number of subarray whose        // maximum element is less than equal to k.        s += ((count * (count + 1)) / 2);    }     return (n * (n + 1) / 2 - s);} // Driven Programint main(){    int arr[] = { 1, 2, 3 };    int k = 2;    int n = sizeof(arr) / sizeof(arr[0]);    cout << countSubarray(arr, n, k);    return 0;}

## Java

 // Java program to count number of subarrays// whose maximum element is greater than K.import java.util.*; class GFG {     // Return number of subarrays whose maximum    // element is less than or equal to K.    static int countSubarray(int arr[], int n, int k)    {         // To store count of subarrays with all        // elements less than or equal to k.        int s = 0;         // Traversing the array.        int i = 0;        while (i < n) {             // If element is greater than k, ignore.            if (arr[i] > k) {                i++;                continue;            }             // Counting the subarray length whose            // each element is less than equal to k.            int count = 0;            while (i < n && arr[i] <= k) {                i++;                count++;            }             // Suming number of subarray whose            // maximum element is less than equal to k.            s += ((count * (count + 1)) / 2);        }         return (n * (n + 1) / 2 - s);    }     // Driver code    public static void main(String[] args)    {         int arr[] = { 1, 2, 3 };        int k = 2;        int n = arr.length;        System.out.print(countSubarray(arr, n, k));    }} // This code is contributed by Anant Agarwal.

## Python3

 # Python program to count# number of subarrays# whose maximum element# is greater than K. # Return number of# subarrays whose maximum# element is less than or equal to K.def countSubarray(arr, n, k):     # To store count of    # subarrays with all    # elements less than    # or equal to k.    s = 0      # Traversing the array.    i = 0    while (i < n):             # If element is greater        # than k, ignore.        if (arr[i] > k):                     i = i + 1            continue                 # Counting the subarray        # length whose        # each element is less        # than equal to k.        count = 0        while (i < n and arr[i] <= k):                     i = i + 1            count = count + 1                   # Suming number of subarray whose        # maximum element is less        # than equal to k.        s = s + ((count*(count + 1))//2)           return (n*(n + 1)//2 - s)     # Driver code arr = [1, 2, 3]k = 2n = len(arr) print(countSubarray(arr, n, k)) # This code is contributed# by Anant Agarwal.

## C#

 // C# program to count number of subarrays// whose maximum element is greater than K.using System; class GFG {     // Return number of subarrays whose maximum    // element is less than or equal to K.    static int countSubarray(int[] arr, int n, int k)    {        // To store count of subarrays with all        // elements less than or equal to k.        int s = 0;         // Traversing the array.        int i = 0;        while (i < n) {             // If element is greater than k, ignore.            if (arr[i] > k) {                i++;                continue;            }             // Counting the subarray length whose            // each element is less than equal to k.            int count = 0;            while (i < n && arr[i] <= k) {                i++;                count++;            }             // Suming number of subarray whose            // maximum element is less than equal to k.            s += ((count * (count + 1)) / 2);        }         return (n * (n + 1) / 2 - s);    }     // Driver code    public static void Main()    {        int[] arr = {1, 2, 3};        int k = 2;        int n = arr.Length;        Console.WriteLine(countSubarray(arr, n, k));    }} // This code is contributed by vt_m.

## PHP

 \$k) {            \$i++;            continue;        }         // Counting the subarray length        // whose each element is less        // than equal to k.        \$count = 0;        while (\$i < \$n and \$arr[\$i] <= \$k) {            \$i++;            \$count++;        }         // Suming number of subarray whose        // maximum element is less than        // equal to k.        \$s += ((\$count * (\$count + 1)) / 2);    }     return (\$n * (\$n + 1) / 2 - \$s);} // Driven Program    \$arr = array( 1, 2, 3 );    \$k = 2;    \$n = count(\$arr);    echo countSubarray(\$arr, \$n, \$k); // This code is contributed by anuj_67.?>

## Javascript



Output:

3

Time Complexity: O(n).
This article is contributed by Anuj Chauhan. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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