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

• Difficulty Level : Medium
• Last Updated : 03 Mar, 2022

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`

Time Complexity: O(n)
Auxiliary Space: O(1)

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