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 <bits/stdc++.h> ` `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 Program ` `int` `main() ` `{ ` ` ` `int` `arr[] = { 1, 2, 3 }; ` ` ` `int` `k = 2; ` ` ` `int` `n = ` `sizeof` `(arr) / ` `sizeof` `(arr[0]); ` ` ` `cout << countSubarray(arr, n, k); ` ` ` `return` `0; ` `} ` |

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## 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. ` |

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## 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 ` `=` `2` `n ` `=` `len` `(arr) ` ` ` `print` `(countSubarray(arr, n, k)) ` ` ` `# This code is contributed ` `# by Anant Agarwal. ` |

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## 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. ` |

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## PHP

`<?php ` `// PHP 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. ` `function` `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` `) { ` ` ` `$i` `++; ` ` ` `continue` `; ` ` ` `} ` ` ` ` ` `// Counting the subarray length ` ` ` `// whose each element is less ` ` ` `// than equal to k. ` ` ` `$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. ` `?> ` |

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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 contribute.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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