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

• Last Updated : 22 Dec, 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 less than or equal to K.
Examples:

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

## Javascript

 ``

Output:

`5`

Time Complexity: O(n), where n is the size of the given array.
Auxiliary Space: O(1), no extra space is required, so it is a constant.

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