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# Maximize sum of averages of subsequences of lengths lying in a given range

• Last Updated : 08 Sep, 2021

Given an array A[] consisting of N integers and two integers X and Y, the task is to find the maximum sum of the averages of each subsequences obtained by splitting the array into subsequences of lengths lying in the range [X, Y].

Note: The values of X and Y are such that it is always possible to make such groups.

Examples:

Input: A[] = {4, 10, 6, 5}, X = 2,  Y = 3
Output: 12.50
Explanation:
Divide the given array into two groups as {4, 10}, {6, 5} such that their sizes lies over the range [2, 3].
The average of the first group = (4 + 10) / 2 = 7.
The average of the second group = (6 + 5) / 2 = 5.5.
Therefore, the sum of average = 7 + 5.5 = 12.5, which is minimum among all possible groups.

Input: A[] = {3, 3, 1}
Output: 3.00

Approach: The given problem can be solved by using the Greedy Approach. The idea is to sort the array in ascending order and choose the groups of size X because the average is inversely proportional to the number of elements. Finally, add the remaining elements to the last group. Follow the below steps to solve the problem:

• Sort the given array arr[] in ascending order.
• Initialize the variables, say sum, res, and count as 0 to store the sum of array elements, result, and the count of elements in the current group.
• Iterate over the range [0, N] using the variable i and perform the following steps:
• Add the value of arr[i] to the variable sum and increase the value of count by 1.
• If the value of count is equal to X, and the remaining array elements can’t form a group then add them to the current group and add the average in the variable res.
• Otherwise, add the average of the group formed so far in the variable res.
• After completing the above steps, print the value of res as the answer.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach``#include ``using` `namespace` `std;` `// Function to find the maximum sum``// of average of groups``void` `maxAverage(``int` `A[], ``int` `N, ``int` `X,``                ``int` `Y)``{``    ``// Sort the given array``    ``sort(A, A + N);` `    ``int` `sum = 0;` `    ``// Stores the sum of averages``    ``double` `res = 0;` `    ``// Stores count of array element``    ``int` `count = 0;` `    ``for` `(``int` `i = 0; i < N; i++) {` `        ``// Add the current value to``        ``// the variable sum``        ``sum += A[i];` `        ``// Increment the count by 1``        ``count++;` `        ``// If the current size is X``        ``if` `(count == X) {` `            ``// If the remaining elements``            ``// can't become a group``            ``if` `(N - i - 1 < X) {``                ``i++;``                ``int` `cnt = 0;` `                ``// Iterate until i is``                ``// less than N``                ``while` `(i < N) {``                    ``cnt++;``                    ``sum += A[i];``                    ``i++;``                ``}` `                ``// Update the value of X``                ``X = X + cnt;` `                ``// Update the average``                ``res += (``double``)sum / ``double``(X);``                ``break``;``            ``}` `            ``// Find the average``            ``res += (``double``)sum / ``double``(X);` `            ``// Reset the sum and count``            ``sum = 0;``            ``count = 0;``        ``}``    ``}` `    ``// Print maximum sum of averages``    ``cout << fixed << setprecision(2)``         ``<< res << ``"\n"``;``}` `// Driver Code``int` `main()``{``    ``int` `A[] = { 4, 10, 6, 5 };``    ``int` `N = ``sizeof``(A) / ``sizeof``(A[0]);``    ``int` `X = 2, Y = 3;` `    ``maxAverage(A, N, X, Y);` `    ``return` `0;``}`

## Java

 `// Java program for the above approach``import` `java.util.*;` `class` `GFG{` `// Function to find the maximum sum``// of average of groups``static` `void` `maxAverage(``int` `A[], ``int` `N, ``int` `X,``                       ``int` `Y)``{``    ` `    ``// Sort the given array``    ``Arrays.sort(A);` `    ``int` `sum = ``0``;` `    ``// Stores the sum of averages``    ``double` `res = ``0``;` `    ``// Stores count of array element``    ``int` `count = ``0``;` `    ``for``(``int` `i = ``0``; i < N; i++)``    ``{``        ` `        ``// Add the current value to``        ``// the variable sum``        ``sum += A[i];` `        ``// Increment the count by 1``        ``count++;` `        ``// If the current size is X``        ``if` `(count == X)``        ``{``            ` `            ``// If the remaining elements``            ``// can't become a group``            ``if` `(N - i - ``1` `< X)``            ``{``                ``i++;``                ``int` `cnt = ``0``;` `                ``// Iterate until i is``                ``// less than N``                ``while` `(i < N)``                ``{``                    ``cnt++;``                    ``sum += A[i];``                    ``i++;``                ``}` `                ``// Update the value of X``                ``X = X + cnt;` `                ``// Update the average``                ``res += (``double``)sum / (``double``)(X);``                ``break``;``            ``}` `            ``// Find the average``            ``res += (``double``)sum / (``double``)(X);` `            ``// Reset the sum and count``            ``sum = ``0``;``            ``count = ``0``;``        ``}``    ``}` `    ``// Print maximum sum of averages``    ``System.out.println(res);``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``int` `A[] = { ``4``, ``10``, ``6``, ``5` `};``    ``int` `N = A.length;``    ``int` `X = ``2``, Y = ``3``;` `    ``maxAverage(A, N, X, Y);``}``}` `// This code is contributed by gauravrajput1`

## Python3

 `# Python 3 program for the above approach` `# Function to find the maximum sum``# of average of groups``def` `maxAverage(A,N,X,Y):``    ``# Sort the given array``    ``A.sort()` `    ``sum` `=` `0` `    ``# Stores the sum of averages``    ``res ``=` `0` `    ``# Stores count of array element``    ``count ``=` `0` `    ``for` `i ``in` `range``(N):``        ``# Add the current value to``        ``# the variable sum``        ``sum` `+``=` `A[i]` `        ``# Increment the count by 1``        ``count ``+``=` `1` `        ``# If the current size is X``        ``if` `(count ``=``=` `X):` `            ``# If the remaining elements``            ``# can't become a group``            ``if` `(N ``-` `i ``-` `1` `< X):``                ``i ``+``=` `1``                ``cnt ``=` `0` `                ``# Iterate until i is``                ``# less than N``                ``while` `(i < N):``                    ``cnt ``+``=` `1``                    ``sum` `+``=` `A[i]``                    ``i ``+``=` `1` `                ``# Update the value of X``                ``X ``=` `X ``+` `cnt` `                ``# Update the average``                ``res ``+``=` `sum` `/` `X``                ``break` `            ``# Find the average``            ``res ``+``=` `sum` `/` `X` `            ``# Reset the sum and count``            ``sum` `=` `0``            ``count ``=` `0` `    ``# Print maximum sum of averages``    ``print``(res)` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ``A ``=` `[``4``, ``10``, ``6``, ``5``]``    ``N ``=` `len``(A)``    ``X ``=` `2``    ``Y ``=` `3` `    ``maxAverage(A, N, X, Y)` `    ``# This code is contributed by ipg2016107.`

## C#

 `// C# program for the above approach``using` `System;` `public` `class` `GFG{` `// Function to find the maximum sum``// of average of groups``static` `void` `maxAverage(``int` `[]A, ``int` `N, ``int` `X,``                       ``int` `Y)``{``    ` `    ``// Sort the given array``    ``Array.Sort(A);` `    ``int` `sum = 0;` `    ``// Stores the sum of averages``    ``double` `res = 0;` `    ``// Stores count of array element``    ``int` `count = 0;` `    ``for``(``int` `i = 0; i < N; i++)``    ``{``        ` `        ``// Add the current value to``        ``// the variable sum``        ``sum += A[i];` `        ``// Increment the count by 1``        ``count++;` `        ``// If the current size is X``        ``if` `(count == X)``        ``{``            ` `            ``// If the remaining elements``            ``// can't become a group``            ``if` `(N - i - 1 < X)``            ``{``                ``i++;``                ``int` `cnt = 0;` `                ``// Iterate until i is``                ``// less than N``                ``while` `(i < N)``                ``{``                    ``cnt++;``                    ``sum += A[i];``                    ``i++;``                ``}` `                ``// Update the value of X``                ``X = X + cnt;` `                ``// Update the average``                ``res += (``double``)sum / (``double``)(X);``                ``break``;``            ``}` `            ``// Find the average``            ``res += (``double``)sum / (``double``)(X);` `            ``// Reset the sum and count``            ``sum = 0;``            ``count = 0;``        ``}``    ``}` `    ``// Print maximum sum of averages``    ``Console.WriteLine(res);``}` `// Driver Code``public` `static` `void` `Main(String[] args)``{``    ``int` `[]A = { 4, 10, 6, 5 };``    ``int` `N = A.Length;``    ``int` `X = 2, Y = 3;` `    ``maxAverage(A, N, X, Y);``}``}` ` `  `// This code is contributed by 29AjayKumar`

## Javascript

 ``
Output:
`12.50`

Time Complexity: O(N * log N)
Auxiliary Space: O(1)

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