Number of subarrays with at-least K size and elements not greater than M

Given an array of N elements, K is the minimum size of the sub-array, and M is the limit of every element in the sub-array, the task is to count the number of sub-arrays with a minimum size of K and all the elements are lesser than or equal to M.

Examples:

Input: arr[] = {-5, 0, -10}, K = 1, M = 15
Output: 6
Explanation: The sub-arrays [-5], [0], [-10], [-5, 0], [0, -10] and [-5, 0, -10] have all elements <= 15 and size>=1.

Input: arr[] = {0, 3, -2, 5, -4, -4}, K = 2, M = 3
Output: 4
Explanation: The sub-arrays are [0, 3], [3,-2], [-4, -4], and [0, 3, -2] have a size of >= K and all elements <=M.

Naive approach: The basic way to solve the problem is as follows:

The idea is to iterate over all the subarrays using nested loops and then check for given conditions if it is satisfied then add it to the answer.

Time Complexity: O(N^2), where N is the size of the array.
Auxiliary Space: O(1)

Efficient Approach: To solve the problem follow the below steps:

• The number of total subarrays of length N is N*(N+1)/2.
• Calculate the maximum length of the array such that all elements are less than equal to M.
• The task is to count the number of subarrays of size at least K out of length we have found.
• Let cnt be the number of consecutive elements counts where all elements are lesser or equal to M.
• The problem is to count the number of subarrays of size at least K out of cnt, we can modify the formulae to be as follows:

(cnt - K + 1)*((cnt - K + 1) + 1)/2

Below is the implementation of the above approach:

// C++ code for the above approach:
#include <bits/stdc++.h>
using namespace std;

// Function to calculate N*(N+1)/2
long countSubarrys(long n) { return n * (n + 1) / 2; }

// Function to count the number of sub-arrays
// with atleast K size and elements
// not greater than M.
long countSubarrays(int a[], int N, int K, int M)
{
    long res = 0;

    // counts curr elements which are <= M
    long cnt = 0;

    for (int i = 0; i < N; i++) {

        // Go on counting the valid elements
        if (a[i] <= M) {
            cnt += 1;
        }

        // As the current ele is > M
        else {

            // Next condition is the
            // sub-array should be >= K
            if (cnt >= K) {

                // Next condition is the
                // sub-array should be >= K

                // Formula to get the
                // number of sub-arrays
                res += countSubarrys(cnt - K + 1);
            }
            cnt = 0;
        }
    }
    // Check our cnt is valid and count the
    // sub-arrays if cnt >= K
    if (cnt >= K) {
        res += countSubarrys(cnt - K + 1);
    }

    return res;
}

// Drivers code
int main()
{
    int a[] = { 2, 0, 11, 3, 0 };
    int N = sizeof(a) / sizeof(a[0]);
    int K = 2, M = 3;

    // Function call
    cout << countSubarrays(a, N, K, M);
    return 0;
}

class GFG {
    // function to calculate N*(N+1)/2
    static long countSubarrys(long n) { 
        return n * (n + 1) / 2;
    }

    // function to count the number of sub-arrays with
    // atleast K size and elements not greater than M.
    static long countSubarrays(int a[], int N, int K, int M) {
        long res = 0;
        long cnt = 0; // counts curr elements which are <= M

        for (int i = 0; i < N; i++) {
            if (a[i] <= M) {  
                cnt += 1; // go on counting the valid elements
            } else { // As the current ele is > M
                if (cnt >= K) { // next condition is the sub-array should be >= K
                    res += countSubarrys(cnt - K + 1); // formula to get the number of sub-arrays
                }
                cnt = 0;
            }
        }
        // check our cnt is valid and count the sub-arrays
        if (cnt >= K) {
            res += countSubarrys(cnt - K + 1);
        }

        return res;
    }

    public static void main(String[] args) {
        int[] a = { 2, 0, 11, 3, 0 };
        int N = a.length;
        int K = 2, M = 3;
        System.out.println(countSubarrays(a, N, K, M));
    }
}

# function to calculate N*(N+1)/2
def countSubarrys(n):
    return n * (n + 1) // 2

# function to count the number of sub-arrays with
# atleast K size and elements not greater than M.
def countSubarrays(a, N, K, M):
    res = 0
    cnt = 0

    for i in range(N):
        if a[i] <= M:
            cnt += 1
        else:
            if cnt >= K:
                res += countSubarrys(cnt - K + 1)
            cnt = 0

    if cnt >= K:
        res += countSubarrys(cnt - K + 1)

    return res

# Driver code
a = [2, 0, 11, 3, 0]
N = len(a)
K = 2
M = 3
print(countSubarrays(a, N, K, M))

using System;

class Program
{
    // Function to calculate N*(N+1)/2
    static long CountSubarrays(long n) => n * (n + 1) / 2;

    // Function to count the number of sub-arrays
    // with at least K size and elements
    // not greater than M.
    static long CountSubarrays(int[] a, int N, int K, int M)
    {
        long res = 0;

        // Count current elements which are <= M
        long cnt = 0;

        for (int i = 0; i < N; i++)
        {
            // Go on counting the valid elements
            if (a[i] <= M)
            {
                cnt += 1;
            }

            // As the current ele is > M
            else
            {
                // Next condition is the sub-array should be >= K
                if (cnt >= K)
                {
                    // Next condition is the sub-array should be >= K

                    // Formula to get the number of sub-arrays
                    res += CountSubarrays(cnt - K + 1);
                }
                cnt = 0;
            }
        }
        // Check if cnt is valid and count the sub-arrays if cnt >= K
        if (cnt >= K)
        {
            res += CountSubarrays(cnt - K + 1);
        }

        return res;
    }

    // Main method
    static void Main()
    {
        int[] a = { 2, 0, 11, 3, 0 };
        int N = a.Length;
        int K = 2, M = 3;

        // Function call
        Console.WriteLine(CountSubarrays(a, N, K, M));
    }
}

// function to calculate N*(N+1)/2
function countSubarrys(n) {
    return (n * (n + 1)) / 2;
}

// function to count the number of sub-arrays with
// atleast K size and elements not greater than M.
function countSubarrays(a, N, K, M) {
    let res = 0;
    let cnt = 0;

    for (let i = 0; i < N; i++) {
        if (a[i] <= M) {
            cnt += 1;
        } else {
            if (cnt >= K) {
                res += countSubarrys(cnt - K + 1);
            }
            cnt = 0;
        }
    }

    if (cnt >= K) {
        res += countSubarrys(cnt - K + 1);
    }

    return res;
}

// Driver code
var a = [2, 0, 11, 3, 0];
var N = a.length;
var K = 2;
var M = 3;
document.write(countSubarrays(a, N, K, M));

2

Time Complexity: O(N), as we are using a loop to traverse N elements.
Auxiliary Space: O(1), as we are not using any extra space.