# Number of non-decreasing sub-arrays of length greater than or equal to K

• Last Updated : 08 Mar, 2022

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 greater than or equal to K.
Examples:

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

Naive approach: A simple approach is to generate all the sub-arrays of length greater 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 + 1) and (X * (X + 1)) / 2 will be added to 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 + 1 = X.
• Number of such sub-arrays starting from the second element = L – K = X – 1.
• Number of such sub-arrays starting from the third element = L – K – 1 = X – 2.
• And so on until 0 i.e. 1 + 2 + 3 + .. + X = (X * (X + 1)) / 2.

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 + 1);` `        ``// Update ret``        ``ret += (x * (x + 1)) / 2;` `        ``// Update i``        ``i = j;``    ``}` `    ``// Return ret``    ``return` `ret;``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 5, 4, 3, 2, 1 };``    ``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 + ``1``);` `        ``// Update ret``        ``ret += (x * (x + ``1``)) / ``2``;` `        ``// Update i``        ``i = j;``    ``}` `    ``// Return ret``    ``return` `ret;``}` `// Driver code``public` `static` `void` `main(String []args)``{``    ``int` `arr[] = { ``5``, ``4``, ``3``, ``2``, ``1` `};``    ``int` `n = arr.length;``    ``int` `k = ``2``;` `    ``System.out.println(findCnt(arr, n, k));``}``}` `// This code is contributed by Rajput-Ji`

## 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 ``+``=` `(x ``*` `(x ``+` `1``)) ``/` `2``;` `        ``# Update i``        ``i ``=` `j;` `    ``# Return ret``    ``return` `ret;` `# Driver code``if` `__name__ ``=``=` `"__main__"` `:` `    ``arr ``=` `[ ``5``, ``4``, ``3``, ``2``, ``1` `];``    ``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 + 1);` `        ``// Update ret``        ``ret += (x * (x + 1)) / 2;` `        ``// Update i``        ``i = j;``    ``}` `    ``// Return ret``    ``return` `ret;``}` `// Driver code``public` `static` `void` `Main(String []args)``{``    ``int` `[]arr = { 5, 4, 3, 2, 1 };``    ``int` `n = arr.Length;``    ``int` `k = 2;` `    ``Console.WriteLine(findCnt(arr, n, k));``}``}` `// This code is contributed by PrinciRaj1992`

## Javascript

 ``
Output:
`0`

Time Complexity: O(n)

Auxiliary Space: O(1)

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