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 approach#include <iostream>using namespace std;// 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; // Stores the sum of subarrays of // size K int sum = 0; // Add the values of first K elements // to the sum for (int i = 0; i < K; i++) { sum += arr[i]; } // Increment the count if the // current subarray is valid if (sum >= K * M) count++; // Traverse the given array for (int i = K; i < N; i++) { // Find the updated sum sum += (arr[i] - arr[i - K]); // Check if current subarray // is valid or not if (sum >= K * M) count++; } // Return the count of subarrays return count;}// Driver Codeint main(){ int arr[] = { 3, 6, 3, 2, 1, 3, 9 }; int K = 2, M = 4; int N = sizeof(arr) / sizeof(arr[0]); cout << countSubArrays(arr, N, K, M); return 0;} |
Java
// Java program for the above approachimport java.util.*;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; // Stores the sum of subarrays of // size K int sum = 0; // Add the values of first K elements // to the sum for (int i = 0; i < K; i++) { sum += arr[i]; } // Increment the count if the // current subarray is valid if (sum >= K * M) count++; // Traverse the given array for (int i = K; i < arr.length; i++) { // Find the updated sum sum += (arr[i] - arr[i - K]); // Check if current subarray // is valid or not if (sum >= K * M) count++; } // Return the count of subarrays return count; }}// This code is contributed by Kdheeraj. |
Python3
# Python 3 code for the above approach# 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 = 0 # Stores the sum of subarrays of # size K sum = 0 # Add the values of first K elements # to the sum for i in range(K): sum += arr[i] # Increment the count if the # current subarray is valid if sum >= K*M: count += 1 # Traverse the given array for i in range(K, N): # Find the updated sum sum += (arr[i] - arr[i - K]) # Check if current subarray # is valid or not if sum >= K*M: count += 1 # Return the count of subarrays return count# Driver Codeif __name__ == '__main__': arr = [3, 6, 3, 2, 1, 3, 9] K = 2 M = 4 N = len(arr) count = countSubArrays(arr, N, K, M) print(count) # This code is contributed by Kdheeraj. |
C#
// C# program for the above approachusing System;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; // Stores the sum of subarrays of // size K int sum = 0; // Add the values of first K elements // to the sum for (int i = 0; i < K; i++) { sum += arr[i]; } // Increment the count if the // current subarray is valid if (sum >= K * M) count++; // Traverse the given array for (int i = K; i < arr.Length; i++) { // Find the updated sum sum += (arr[i] - arr[i - K]); // Check if current subarray // is valid or not if (sum >= K * M) count++; } // Return the count of subarrays return count; }}// This code is contributed by AnkThon |
Javascript
<script> // JavaScript Program to implement // the above approach // 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; // Stores the sum of subarrays of // size K let sum = 0; // Add the values of first K elements // to the sum for (let i = 0; i < K; i++) { sum += arr[i]; } // Increment the count if the // current subarray is valid if (sum >= K * M) count++; // Traverse the given array for (let i = K; i < N; i++) { // Find the updated sum sum += (arr[i] - arr[i - K]); // Check if current subarray // is valid or not if (sum >= K * M) count++; } // Return the count of subarrays return count; } // Driver Code let arr = [3, 6, 3, 2, 1, 3, 9]; let K = 2, M = 4; let N = arr.length; document.write(countSubArrays(arr, N, K, M)); // This code is contributed by Potta Lokesh </script> |
Output:
3
Time Complexity: O(N)
Auxiliary Space: O(1)
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