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Number of non-decreasing sub-arrays of length less than or equal to K
• Last Updated : 11 Nov, 2019

Given an array arr[] of N elements and an integer K, the task is to find the number of non-decreasing sub-arrays of length less than or equal to K.

Examples:

Input: arr[] = {1, 2, 3}, K = 2
Output: 5
{1}, {2}, {3}, {1, 2} and {2, 3} are the valid subarrays.

Input: arr[] = {3, 2, 1}, K = 1
Output: 3

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Naive approach: A simple approach is to generate all the sub-arrays of length less than or equal to K and then check whether the sub-array satisfies the condition. Thus, the time complexity of the approach will be O(N3).

Efficient approach: A better approach will be using the two-pointer technique.

• For any index i, find the largest index j such that the sub-array arr[i…j] is non-decreasing. This can be achieved by simply increasing the value of j, starting from i + 1 and checking whether arr[j] is greater than arr[j – 1].
• Lets say that the length of the sub-array found in the previous step is L. Calculate X = max(0, L – K) and (L * (L + 1)) / 2 – (X * (X + 1)) / 2 will be added to the final answer. This is because for an array of length L, the number of sub-arrays with length ≥ K.
• Number of such sub-arrays starting from the first element = L – K = X.
• Number of such sub-arrays starting from the second element = L – K – 1 = X – 1.
• Number of such sub-arrays starting from the third element = L – K – 2 = X – 2.
• And so on until 0 i.e. 1 + 2 + 3 + .. + X = (X * (X + 1)) / 2. If this value is subtracted from the total increasing subarrays then the result will be the count of increasing subarrays of length less than or equal to K

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;`` ` `// Function to return the required count``int` `findCnt(``int``* arr, ``int` `n, ``int` `k)``{``    ``// To store the final result``    ``int` `ret = 0;`` ` `    ``// Two pointer loop``    ``int` `i = 0;``    ``while` `(i < n) {`` ` `        ``// Initialising j``        ``int` `j = i + 1;`` ` `        ``// Looping till the subarray increases``        ``while` `(j < n and arr[j] >= arr[j - 1])``            ``j++;``        ``int` `x = max(0, j - i - k);`` ` `        ``// Update ret``        ``ret += ((j - i) * (j - i + 1)) / 2 - (x * (x + 1)) / 2;`` ` `        ``// Update i``        ``i = j;``    ``}`` ` `    ``// Return ret``    ``return` `ret;``}`` ` `// Driver code``int` `main()``{``    ``int` `arr[] = { 1, 2, 3 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(``int``);``    ``int` `k = 2;`` ` `    ``cout << findCnt(arr, n, k);`` ` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``class` `GFG ``{`` ` `// Function to return the required count``static` `int` `findCnt(``int``[] arr, ``int` `n, ``int` `k)``{``    ``// To store the final result``    ``int` `ret = ``0``;`` ` `    ``// Two pointer loop``    ``int` `i = ``0``;``    ``while` `(i < n)``    ``{`` ` `        ``// Initialising j``        ``int` `j = i + ``1``;`` ` `        ``// Looping till the subarray increases``        ``while` `(j < n && arr[j] >= arr[j - ``1``])``            ``j++;``        ``int` `x = Math.max(``0``, j - i - k);`` ` `        ``// Update ret``        ``ret += ((j - i) * (j - i + ``1``)) / ``2` `- ``                          ``(x * (x + ``1``)) / ``2``;`` ` `        ``// Update i``        ``i = j;``    ``}`` ` `    ``// Return ret``    ``return` `ret;``}`` ` `// Driver code``public` `static` `void` `main(String []args)``{``    ``int` `arr[] = { ``1``, ``2``, ``3` `};``    ``int` `n = arr.length;``    ``int` `k = ``2``;`` ` `    ``System.out.println(findCnt(arr, n, k));``}``}`` ` `// This code is contributed by 29AjayKumar`

## Python3

 `# Python3 implementation of the approach `` ` `# Function to return the required count ``def` `findCnt(arr, n, k) :`` ` `    ``# To store the final result ``    ``ret ``=` `0``; `` ` `    ``# Two pointer loop ``    ``i ``=` `0``; ``    ``while` `(i < n) :`` ` `        ``# Initialising j ``        ``j ``=` `i ``+` `1``; `` ` `        ``# Looping till the subarray increases ``        ``while` `(j < n ``and` `arr[j] >``=` `arr[j ``-` `1``]) :``            ``j ``+``=` `1``; ``             ` `        ``x ``=` `max``(``0``, j ``-` `i ``-` `k); `` ` `        ``# Update ret ``        ``ret ``+``=` `((j ``-` `i) ``*` `(j ``-` `i ``+` `1``)) ``/``/` `2` `-` `\``                          ``(x ``*` `(x ``+` `1``)) ``/` `2``; `` ` `        ``# Update i ``        ``i ``=` `j; `` ` `    ``# Return ret ``    ``return` `ret; `` ` `# Driver code ``if` `__name__ ``=``=` `"__main__"` `: `` ` `    ``arr ``=` `[ ``1``, ``2``, ``3` `]; ``    ``n ``=` `len``(arr); ``    ``k ``=` `2``; `` ` `    ``print``(findCnt(arr, n, k)); `` ` `# This code is contributed by AnkitRai01`

## C#

 `// C# implementation of the approach``using` `System;``                     ` `class` `GFG``{`` ` `// Function to return the required count``static` `int` `findCnt(``int``[] arr, ``int` `n, ``int` `k)``{``    ``// To store the final result``    ``int` `ret = 0;`` ` `    ``// Two pointer loop``    ``int` `i = 0;``    ``while` `(i < n)``    ``{`` ` `        ``// Initialising j``        ``int` `j = i + 1;`` ` `        ``// Looping till the subarray increases``        ``while` `(j < n && arr[j] >= arr[j - 1])``            ``j++;``        ``int` `x = Math.Max(0, j - i - k);`` ` `        ``// Update ret``        ``ret += ((j - i) * (j - i + 1)) / 2 - ``                        ``(x * (x + 1)) / 2;`` ` `        ``// Update i``        ``i = j;``    ``}`` ` `    ``// Return ret``    ``return` `ret;``}`` ` `// Driver code``public` `static` `void` `Main(String []args)``{``    ``int` `[]arr = { 1, 2, 3 };``    ``int` `n = arr.Length;``    ``int` `k = 2;`` ` `    ``Console.WriteLine(findCnt(arr, n, k));``}``}`` ` `// This code is contributed by Rajput-Ji`
Output:
```5
``` My Personal Notes arrow_drop_up