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 

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 <bits/stdc++.h>
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

<script>

// Javascript implementation of the approach

// Function to return the required count
function findCnt(arr, n, k)
{
    // To store the final result
    var ret = 0;

    // Two pointer loop
    var i = 0;
    while (i < n) {

        // Initialising j
        var j = i + 1;

        // Looping till the subarray increases
        while (j < n && arr[j] >= arr[j - 1])
            j++;
        var 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
var arr = [5, 4, 3, 2, 1];
var n = arr.length;
var k = 2;
document.write( findCnt(arr, n, k));


</script>

Output:

0

Time Complexity: O(n)

Auxiliary Space: O(1)