Given an array arr[] consisting of N integers and two positive integers K and M, the task is to find the number of subarrays of size K whose average is at least M.
Examples:
Input: arr[] = {2, 3, 3, 4, 4, 4, 5, 6, 6}, K = 3, M = 4
Output: 4
Explanation:
Below are the subarrays of size K(= 3) whose average is at least M(= 4) as:
- arr[3, 5]: The average is 4 which is at least M(= 4).
- arr[4, 6]: The average is 4.33 which is at least M(= 4).
- arr[5, 7]: The average is 5 which is at least M(= 4).
- arr[6, 8]: The average is 5.66 which is at least M(= 4).
Therefore, the count of the subarray is given by 4.
Input: arr[] = {3, 6, 3, 2, 1, 3, 9] K = 2, M = 4
Output: 3
Approach: The given problem can be solved by using the Two Pointers and Sliding Window Technique. Follow the steps below to solve the given problem:
- Initialize a variable, say count as 0 that stores the count of all possible subarrays.
- Initialize a variable, say sum as 0 that stores the sum of elements of the subarray of size K.
- Find the sum of the first K array elements and store it in the variable sum. If the value of sum is at least M*K, then increment the value of count by 1.
- Traverse the given array arr[] over the range [K, N – 1] using the variable i and perform the following steps:
- Add the value of arr[i] to the variable sum and subtract the value of arr[i – K] from the sum.
- If the value of sum is at least M*K, then increment the value of count by 1.
- After completing the above steps, print the value of count as the resultant count of subarrays.
Below is the implementation of the above approach:
C++
// C++ program for the above approachÂ
#include <iostream>using namespace std;Â
// Function to count the subarrays of// size K having average at least Mint countSubArrays(int arr[], int N,                   int K, int M){    // Stores the resultant count of    // subarray    int count = 0;Â
    // Stores the sum of subarrays of    // size K    int sum = 0;Â
    // Add the values of first K elements    // to the sum    for (int i = 0; i < K; i++) {        sum += arr[i];    }Â
    // Increment the count if the    // current subarray is valid    if (sum >= K * M)        count++;Â
    // Traverse the given array    for (int i = K; i < N; i++) {Â
        // Find the updated sum        sum += (arr[i] - arr[i - K]);Â
        // Check if current subarray        // is valid or not        if (sum >= K * M)            count++;    }Â
    // Return the count of subarrays    return count;}Â
// Driver Codeint main(){Â Â Â Â int arr[] = { 3, 6, 3, 2, 1, 3, 9 };Â Â Â Â int K = 2, M = 4;Â Â Â Â int N = sizeof(arr) / sizeof(arr[0]);Â
    cout << countSubArrays(arr, N, K, M);Â
    return 0;} |
Java
// Java program for the above approachimport java.util.*;Â
class GFG {       // Driver Code    public static void main(String[] args)    {        int[] arr = { 3, 6, 3, 2, 1, 3, 9 };        int K = 2, M = 4;        System.out.println(countSubArrays(arr, K, M));    }       // Function to count the subarrays of    // size K having average at least M    public static int countSubArrays(int[] arr, int K,                                     int M)    {               // Stores the resultant count of        // subarray        int count = 0;Â
        // Stores the sum of subarrays of        // size K        int sum = 0;Â
        // Add the values of first K elements        // to the sum        for (int i = 0; i < K; i++) {            sum += arr[i];        }Â
        // Increment the count if the        // current subarray is valid        if (sum >= K * M)            count++;Â
        // Traverse the given array        for (int i = K; i < arr.length; i++) {Â
            // Find the updated sum            sum += (arr[i] - arr[i - K]);Â
            // Check if current subarray            // is valid or not            if (sum >= K * M)                count++;        }Â
        // Return the count of subarrays        return count;    }}Â
// This code is contributed by Kdheeraj. |
Python3
# Python 3 code for the above approachÂ
# Function to count the subarrays of# size K having average at least Mdef countSubArrays(arr, N, K, M):       # Stores the resultant count of    # subarray    count = 0Â
    # Stores the sum of subarrays of    # size K    sum = 0Â
    # Add the values of first K elements    # to the sum    for i in range(K):        sum += arr[i]Â
    # Increment the count if the    # current subarray is valid    if sum >= K*M:        count += 1Â
    # Traverse the given array    for i in range(K, N):Â
        # Find the updated sum        sum += (arr[i] - arr[i - K])Â
        # Check if current subarray        # is valid or not        if sum >= K*M:            count += 1Â
    # Return the count of subarrays    return countÂ
# Driver Codeif __name__ == '__main__':Â Â Â Â arr = [3, 6, 3, 2, 1, 3, 9]Â Â Â Â K = 2Â Â Â Â M = 4Â Â Â Â N = len(arr)Â Â Â Â count = countSubArrays(arr, N, K, M)Â Â Â Â print(count)Â
    # This code is contributed by Kdheeraj. |
C#
// C# program for the above approachÂ
using System;Â
public class GFG {       // Driver Code    public static void Main(String[] args)    {        int[] arr = { 3, 6, 3, 2, 1, 3, 9 };        int K = 2, M = 4;        Console.WriteLine(countSubArrays(arr, K, M));    }       // Function to count the subarrays of    // size K having average at least M    public static int countSubArrays(int[] arr, int K,                                     int M)    {               // Stores the resultant count of        // subarray        int count = 0;Â
        // Stores the sum of subarrays of        // size K        int sum = 0;Â
        // Add the values of first K elements        // to the sum        for (int i = 0; i < K; i++) {            sum += arr[i];        }Â
        // Increment the count if the        // current subarray is valid        if (sum >= K * M)            count++;Â
        // Traverse the given array        for (int i = K; i < arr.Length; i++) {Â
            // Find the updated sum            sum += (arr[i] - arr[i - K]);Â
            // Check if current subarray            // is valid or not            if (sum >= K * M)                count++;        }Â
        // Return the count of subarrays        return count;    }}Â
// This code is contributed by AnkThon |
Javascript
<script>       // JavaScript Program to implement       // the above approachÂ
       // Function to count the subarrays of       // size K having average at least M       function countSubArrays(arr, N,           K, M)        {           // Stores the resultant count of           // subarray           let count = 0;Â
           // Stores the sum of subarrays of           // size K           let sum = 0;Â
           // Add the values of first K elements           // to the sum           for (let i = 0; i < K; i++) {               sum += arr[i];           }Â
           // Increment the count if the           // current subarray is valid           if (sum >= K * M)               count++;Â
           // Traverse the given array           for (let i = K; i < N; i++) {Â
               // Find the updated sum               sum += (arr[i] - arr[i - K]);Â
               // Check if current subarray               // is valid or not               if (sum >= K * M)                   count++;           }Â
           // Return the count of subarrays           return count;       }Â
       // Driver Code       let arr = [3, 6, 3, 2, 1, 3, 9];       let K = 2, M = 4;       let N = arr.length;Â
       document.write(countSubArrays(arr, N, K, M));Â
    // This code is contributed by Potta LokeshÂ
   </script> |
Output:Â
3
Â
Time Complexity: O(N)
Auxiliary Space: O(1)Â
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