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

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: 3
{1, 2}, {2, 3} and {1, 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 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 `

Output:

```0
```

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