大小为 K 且平均至少为 M 的子数组的计数

给定一个由N个整数和两个正整数K和M组成的数组arr[] ，任务是找到平均至少为 M的大小为K的子数组的数量。

例子：

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

编程需要懂一点英语

方法：给定的问题可以通过使用两个指针和滑动窗口技术来解决。请按照以下步骤解决给定的问题：

初始化一个变量，比如count为0 ，它存储所有可能的子数组的计数。
初始化一个变量，比如sum为0 ，它存储大小为K的子数组的元素之和。
找到前K个数组元素的总和并将其存储在变量sum中。如果sum的值至少为 M*K ，则将count的值增加1 。
使用变量i在[K, N – 1]范围内遍历给定数组arr[]并执行以下步骤：
- 将arr[i]的值添加到变量sum并从 sum 中减去arr[i – K]的值。
- 如果sum的值至少为 M*K ，则将count的值增加1 。
完成上述步骤后，将count的值打印为子数组的结果计数。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to count the subarrays of
// size K having average at least M
int 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 Code
int 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 approach
import 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 M
def 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 Code
if __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 approach
 
using 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

输出：

时间复杂度： O(N)
辅助空间： O(1)