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Count subarrays consisting of first K natural numbers in descending order
• Difficulty Level : Medium
• Last Updated : 13 Apr, 2021

Given an array arr[] of size N and an integer K, the task is to count the number of subarrays which consists of first K natural numbers in descending order.

Examples:

Input: arr[] = {1, 2, 3, 7, 9, 3, 2, 1, 8, 3, 2, 1}, K = 3
Output: 2
Explanation: The subarray {3, 2, 1} occurs twice in the array.

Input: arr = {100, 7, 6, 5, 4, 3, 2, 1, 100}, K = 6
Output: 1

Approach: The idea is to traverse the array and check if the required decreasing sequence is present starting from the current index or not. Follow the steps below to solve the problem:

• Initialize two variables, temp to K, that checks the pattern, and count with 0, to store the count of total subarray matched.
• Traverse the array arr[] using the variable i and do the following:
• If arr[i] is equal to temp and the value of temp is 1, then increment the count by 1 and update temp as K. Else decrement temp by 1.
• Otherwise, update temp as temp = K and if arr[i] is equal to K, decrement i by 1.
• After the above steps, print the value of count as the result.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach` `#include ``using` `namespace` `std;` `// Function to count subarray having``// the decreasing sequence K to 1``int` `CountSubarray(``int` `arr[], ``int` `n,``                  ``int` `k)``{``    ``int` `temp = k, count = 0;` `    ``// Traverse the array``    ``for` `(``int` `i = 0; i < n; i++) {` `        ``// Check if required sequence``        ``// is present or not``        ``if` `(arr[i] == temp) {``            ``if` `(temp == 1) {``                ``count++;``                ``temp = k;``            ``}``            ``else``                ``temp--;``        ``}` `        ``// Reset temp to k``        ``else` `{``            ``temp = k;``            ``if` `(arr[i] == k)``                ``i--;``        ``}``    ``}` `    ``// Return the count``    ``return` `count;``}` `// Driver Code``int` `main()``{``    ``int` `arr[] = { 1, 2, 3, 7, 9, 3,``                  ``2, 1, 8, 3, 2, 1 };``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr);``    ``int` `K = 3;` `    ``// Function Call``    ``cout << CountSubarray(arr, N, K);` `    ``return` `0;``}` `// This code is contributed by Dharanendra L V`

## Java

 `// Java program for the above approach``import` `java.util.*;``class` `GFG``{` `  ``// Function to count subarray having``  ``// the decreasing sequence K to 1``  ``static` `int` `CountSubarray(``int` `arr[], ``int` `n,``                           ``int` `k)``  ``{``    ``int` `temp = k, count = ``0``;` `    ``// Traverse the array``    ``for` `(``int` `i = ``0``; i < n; i++) {` `      ``// Check if required sequence``      ``// is present or not``      ``if` `(arr[i] == temp) {``        ``if` `(temp == ``1``) {``          ``count++;``          ``temp = k;``        ``}``        ``else``          ``temp--;``      ``}` `      ``// Reset temp to k``      ``else` `{``        ``temp = k;``        ``if` `(arr[i] == k)``          ``i--;``      ``}``    ``}` `    ``// Return the count``    ``return` `count;``  ``}` `  ``// Driver code``  ``public` `static` `void` `main(String[] args)``  ``{``    ``int` `arr[] = { ``1``, ``2``, ``3``, ``7``, ``9``, ``3``,``                 ``2``, ``1``, ``8``, ``3``, ``2``, ``1` `};``    ``int` `N = arr.length;``    ``int` `K = ``3``;` `    ``// Function Call``    ``System.out.println(CountSubarray(arr, N, K));``  ``}``}` `// This code is contributed by shivanisinghss2110`

## Python3

 `# Python3 program for the above approach` `# Function to count subarray having``# the decreasing sequence K to 1``def` `CountSubarray(arr, n, k):``    ` `    ``temp ``=` `k``    ``count ``=` `0` `    ``# Traverse the array``    ``for` `i ``in` `range``(n):` `        ``# Check if required sequence``        ``# is present or not``        ``if` `(arr[i] ``=``=` `temp):``            ``if` `(temp ``=``=` `1``):``                ``count ``+``=` `1``                ``temp ``=` `k``            ``else``:``                   ``temp ``-``=` `1` `        ``# Reset temp to k``        ``else``:``            ``temp ``=` `k``            ` `            ``if` `(arr[i] ``=``=` `k):``                ``i ``-``=` `1` `    ``# Return the count``    ``return` `count` `# Driver Code``if` `__name__ ``=``=` `"__main__"``:` `    ``arr ``=` `[ ``1``, ``2``, ``3``, ``7``, ``9``, ``3``,``            ``2``, ``1``, ``8``, ``3``, ``2``, ``1` `]``    ``N ``=` `len``(arr)``    ``K ``=` `3` `    ``# Function Call``    ``print``(CountSubarray(arr, N, K))` `# This code is contributed by chitranayal`

## C#

 `// C# program for the above approach``using` `System;` `class` `GFG{` `// Function to count subarray having``// the decreasing sequence K to 1``static` `int` `CountSubarray(``int``[] arr,``                         ``int` `n, ``int` `k)``{``    ``int` `temp = k, count = 0;` `    ``// Traverse the array``    ``for``(``int` `i = 0; i < n; i++)``    ``{``        ` `        ``// Check if required sequence``        ``// is present or not``        ``if` `(arr[i] == temp)``        ``{``            ``if` `(temp == 1)``            ``{``                ``count++;``                ``temp = k;``            ``}``            ``else``                ``temp--;``        ``}` `        ``// Reset temp to k``        ``else``        ``{``            ``temp = k;``            ` `            ``if` `(arr[i] == k)``                ``i--;``        ``}``    ``}` `    ``// Return the count``    ``return` `count;``}` `// Driver code``static` `public` `void` `Main()``{``    ``int``[] arr = { 1, 2, 3, 7, 9, 3,``                  ``2, 1, 8, 3, 2, 1 };``    ``int` `N = arr.Length;``    ``int` `K = 3;` `    ``// Function Call``    ``Console.Write(CountSubarray(arr, N, K));``}``}` `// This code is contributed by Dharanendra L V`

## Javascript

 ``
Output:
`2`

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

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