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# Number of non-decreasing sub-arrays of length K

• Difficulty Level : Medium
• Last Updated : 12 May, 2021

Given an array arr[] of length N, the task is to find the number of non-decreasing sub-arrays of length K.
Examples:

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

Naive approach Generate all the sub-arrays of length K and then check whether the sub-array satisfies the condition. Thus, the time complexity of the approach will be O(N * K).
Better approach: A better approach will be using two-pointer technique. Let’s say the current index is i

• Find the largest index j, such that the sub-array arr[i…j] is non-decreasing. This can be achieved by simply incrementing the value of j starting from i + 1 and checking whether arr[j] is greater than arr[j – 1].
• Let’s say the length of the sub-array found in the previous step is L. The number of sub-arrays of length K contained in it will be max(L – K + 1, 0).
• Now, update i = j and repeat the above steps while i is in the index range.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Function to return the count of``// increasing subarrays of length k``int` `cntSubArrays(``int``* arr, ``int` `n, ``int` `k)``{``    ``// To store the final result``    ``int` `res = 0;` `    ``int` `i = 0;``    ``// Two pointer loop``    ``while` `(i < n) {` `        ``// Initialising j``        ``int` `j = i + 1;` `        ``// Looping till the subarray increases``        ``while` `(j < n and arr[j] >= arr[j - 1])``            ``j++;` `        ``// Updating the required count``        ``res += max(j - i - k + 1, 0);` `        ``// Updating i``        ``i = j;``    ``}` `    ``// Returning res``    ``return` `res;``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 1, 2, 3, 2, 5 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(``int``);``    ``int` `k = 2;` `    ``cout << cntSubArrays(arr, n, k);` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``class` `GFG``{` `// Function to return the count of``// increasing subarrays of length k``static` `int` `cntSubArrays(``int` `[]arr, ``int` `n, ``int` `k)``{``    ``// To store the final result``    ``int` `res = ``0``;` `    ``int` `i = ``0``;``    ` `    ``// Two pointer loop``    ``while` `(i < n)``    ``{` `        ``// Initialising j``        ``int` `j = i + ``1``;` `        ``// Looping till the subarray increases``        ``while` `(j < n && arr[j] >= arr[j - ``1``])``            ``j++;` `        ``// Updating the required count``        ``res += Math.max(j - i - k + ``1``, ``0``);` `        ``// Updating i``        ``i = j;``    ``}` `    ``// Returning res``    ``return` `res;``}` `// Driver code``public` `static` `void` `main(String []args)``{``    ``int` `arr[] = { ``1``, ``2``, ``3``, ``2``, ``5` `};``    ``int` `n = arr.length;``    ``int` `k = ``2``;` `    ``System.out.println(cntSubArrays(arr, n, k));``}``}` `// This code is contributed by PrinciRaj1992`

## Python3

 `# Python3 implementation of the approach` `# Function to return the count of``# increasing subarrays of length k``def` `cntSubArrays(arr, n, k) :` `    ``# To store the final result``    ``res ``=` `0``;` `    ``i ``=` `0``;``    ` `    ``# Two pointer loop``    ``while` `(i < n) :` `        ``# Initialising j``        ``j ``=` `i ``+` `1``;` `        ``# Looping till the subarray increases``        ``while` `(j < n ``and` `arr[j] >``=` `arr[j ``-` `1``]) :``            ``j ``+``=` `1``;` `        ``# Updating the required count``        ``res ``+``=` `max``(j ``-` `i ``-` `k ``+` `1``, ``0``);` `        ``# Updating i``        ``i ``=` `j;` `    ``# Returning res``    ``return` `res;` `# Driver code``if` `__name__ ``=``=` `"__main__"` `:``    ` `    ``arr ``=` `[ ``1``, ``2``, ``3``, ``2``, ``5` `];``    ``n ``=` `len``(arr);``    ``k ``=` `2``;` `    ``print``(cntSubArrays(arr, n, k));``    ` `# This code is contributed by AnkitRai01`

## C#

 `// C# implementation of the approach``using` `System;``    ` `class` `GFG``{` `// Function to return the count of``// increasing subarrays of length k``static` `int` `cntSubArrays(``int` `[]arr, ``int` `n, ``int` `k)``{``    ``// To store the final result``    ``int` `res = 0;` `    ``int` `i = 0;``    ` `    ``// Two pointer loop``    ``while` `(i < n)``    ``{` `        ``// Initialising j``        ``int` `j = i + 1;` `        ``// Looping till the subarray increases``        ``while` `(j < n && arr[j] >= arr[j - 1])``            ``j++;` `        ``// Updating the required count``        ``res += Math.Max(j - i - k + 1, 0);` `        ``// Updating i``        ``i = j;``    ``}` `    ``// Returning res``    ``return` `res;``}` `// Driver code``public` `static` `void` `Main(String []args)``{``    ``int` `[]arr = { 1, 2, 3, 2, 5 };``    ``int` `n = arr.Length;``    ``int` `k = 2;` `    ``Console.WriteLine(cntSubArrays(arr, n, k));``}``}` `// This code is contributed by Rajput-Ji`

## Javascript

 ``
Output:
`3`

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