至少X和最大Y大小的子数组的最大平均值

给定一个数组arr []和两个整数X和Y。任务是找到一个大小至少为X和最大Y为平均值(最大的子数组)的子数组。
例子：

Input: arr[] = {1, 2, 3, 4, 5} X = 2, Y = 3 
Output: 4.5 
We can take the sub-array {4, 5} which gives us the maximum average. 
Input: arr[] = {6, 7, 8, 3, 2, 4, 2} X = 2, Y = 4 
Output: 7.5 
 

方法：对从X到Y大小的每个大小的子数组进行迭代，并找到所有此类子数组中的最大平均值。我们可以使用两个嵌套的for循环来迭代大小从X到Y的所有子数组。为了减少时间复杂度，我们可以使用前缀求和数组来获取O(1)复杂度的任何子数组的和。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to return the maximum average
// of the sub-array with size
// atleast x and atmost y
double maxAvgSubArray(int a[], int n, int x, int y)
{
 
    // Calculate the prefix sum array
    int prefix[n];
    prefix[0] = a[0];
    for (int i = 1; i < n; i++)
        prefix[i] = prefix[i - 1] + a[i];
 
    double maximum = 0;
 
    // Iterate over all sub-arrays
    for (int i = 0; i < n; i++) {
 
        // Sub-arrays of size X to Y
        for (int j = i + x - 1; j < i + y && j < n; j++) {
 
            // Get the sum of the sub-array
            double sum = prefix[j];
            if (i > 0)
                sum -= prefix[i - 1];
 
            // Find average of sub-array
            double current = sum / (double)(j - i + 1);
 
            // Store the maximum of average
            maximum = max(maximum, current);
        }
    }
 
    return maximum;
}
 
// Driver code
int main()
{
    int a[] = { 6, 7, 8, 3, 2, 4, 2 };
    int X = 2, Y = 4;
    int n = sizeof(a) / sizeof(a[0]);
    cout << maxAvgSubArray(a, n, X, Y);
 
    return 0;
}

Java

// Java implementation of the approach
class GfG {
 
    // Function to return the maximum average
    // of the sub-array with size
    // atleast x and atmost y
    static double maxAvgSubArray(int a[], int n, int x, int y)
    {
 
        // Calculate the prefix sum array
        int prefix[] = new int[n];
        prefix[0] = a[0];
        for (int i = 1; i < n; i++)
            prefix[i] = prefix[i - 1] + a[i];
 
        double maximum = 0;
 
        // Iterate over all sub-arrays
        for (int i = 0; i < n; i++) {
 
            // Sub-arrays of size X to Y
            for (int j = i + x - 1; j < i + y && j < n; j++) {
 
                // Get the sum of the sub-array
                double sum = prefix[j];
                if (i > 0)
                    sum -= prefix[i - 1];
 
                // Find average of sub-array
                double current = sum / (double)(j - i + 1);
 
                // Store the maximum of average
                maximum = Math.max(maximum, current);
            }
        }
 
        return maximum;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int a[] = { 6, 7, 8, 3, 2, 4, 2 };
        int X = 2, Y = 4;
        int n = a.length;
        System.out.println(maxAvgSubArray(a, n, X, Y));
    }
}

Python3

# Python3 implementation of the approach
 
# Function to return the maximum average
# of the sub-array with size
# atleast x and atmost y
def maxAvgSubArray(a, n, x, y) :
 
    # Calculate the prefix sum array
    prefix = [0] * n ;
    prefix[0] = a[0];
    for i in range(1, n) :
        prefix[i] = prefix[i - 1] + a[i];
         
    maximum = 0;
     
    # Iterate over all sub-arrays
    for i in range(n) :
        j = i + x - 1
         
        # Sub-arrays of size X to Y
        while(j < i + y and j < n) :
             
            # Get the sum of the sub-array
            sum = prefix[j];
             
            if (i > 0) :
                sum -= prefix[i - 1];
             
            # Find average of sub-array
            current = sum / (j - i + 1);
         
            # Store the maximum of average
            maximum = max(maximum, current);
             
            j += 1
    return maximum;
 
# Driver code
if __name__ == "__main__" :
 
    a = [ 6, 7, 8, 3, 2, 4, 2 ];
    X = 2; Y = 4;
     
    n = len(a);
    print(maxAvgSubArray(a, n, X, Y));
 
# This code is contributed by Ryuga

C#

// C# implementation of the approach
using System;
 
class GFG
{
 
    // Function to return the maximum
    // average of the sub-array with
    // size atleast x and atmost y
    public static double maxAvgSubArray(int[] a, int n,
                                        int x, int y)
    {
 
        // Calculate the prefix sum array
        int[] prefix = new int[n];
        prefix[0] = a[0];
        for (int i = 1; i < n; i++)
        {
            prefix[i] = prefix[i - 1] + a[i];
        }
 
        double maximum = 0;
 
        // Iterate over all sub-arrays
        for (int i = 0; i < n; i++)
        {
 
            // Sub-arrays of size X to Y
            for (int j = i + x - 1;
                     j < i + y && j < n; j++)
            {
 
                // Get the sum of the sub-array
                double sum = prefix[j];
                if (i > 0)
                {
                    sum -= prefix[i - 1];
                }
 
                // Find average of sub-array
                double current = sum / (double)(j - i + 1);
 
                // Store the maximum of average
                maximum = Math.Max(maximum, current);
            }
        }
 
        return maximum;
    }
 
    // Driver code
    public static void Main(string[] args)
    {
        int[] a = new int[] {6, 7, 8, 3, 2, 4, 2};
        int X = 2, Y = 4;
        int n = a.Length;
        Console.WriteLine(maxAvgSubArray(a, n, X, Y));
    }
}
 
// This code is contributed by Shrikant13

PHP

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                $sum -= $prefix[$i - 1];
 
            // Find average of sub-array
            $current = ($sum / ($j - $i + 1));
 
            // Store the maximum of average
            $maximum = max($maximum, $current);
        }
    }
 
    return $maximum;
}
 
// Driver code
$a = array(6, 7, 8, 3, 2, 4, 2);
$X = 2; $Y = 4;
$n = sizeof($a);
echo maxAvgSubArray($a, $n, $X, $Y);
 
// This code is contributed by Akanksha Rai
?>

Javascript

 
// javascript program for the above approach
 
    // Function to return the maximum average
    // of the sub-array with size
    // atleast x and atmost y
    function maxAvgSubArray(a, n, x, y)
    {
 
        // Calculate the prefix sum array
        let prefix = new Array(n).fill(0);
        prefix[0] = a[0];
        for (let i = 1; i < n; i++)
            prefix[i] = prefix[i - 1] + a[i];
 
        let maximum = 0;
 
        // Iterate over all sub-arrays
        for (let i = 0; i < n; i++) {
 
            // Sub-arrays of size X to Y
            for (let j = i + x - 1; j < i + y && j < n; j++) {
 
                // Get the sum of the sub-array
                let sum = prefix[j];
                if (i > 0)
                    sum -= prefix[i - 1];
 
                // Find average of sub-array
                let current = sum / (j - i + 1);
 
                // Store the maximum of average
                maximum = Math.max(maximum, current);
            }
        }
 
        return maximum;
    }
 
// Driver Code
     
        let a = [ 6, 7, 8, 3, 2, 4, 2 ];
        let X = 2, Y = 4;
        let n = a.length;
        document.write(maxAvgSubArray(a, n, X, Y));
 

输出：

7.5

时间复杂度： O(N *(YX))
辅助空间： O(N)

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