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# Number of subsets whose mean is maximum

• Last Updated : 14 Sep, 2021

Given an array arr[] of size N, the task is to count the number of subsets of arr[] whose mean is maximum.

Examples:

Input: arr[] = {1, 2, 1, 2}
Output:
Subsets with maximum mean are {2}, {2} and {2, 2}.

Input: arr[] = {1}
Output: 1

Approach: The maximum value for the mean of any subset will be when the subset will only consist of the maximum element from the array. So, in order to count all the possible subsets, find the frequency of the maximum element from the array say cnt and the count of possible subsets will be 2cnt – 1.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Function to return the count of``// subsets with the maximum mean``int` `cntSubSets(``int` `arr[], ``int` `n)``{` `    ``// Maximum value from the array``    ``int` `maxVal = *max_element(arr, arr + n);` `    ``// To store the number of times maximum``    ``// element appears in the array``    ``int` `cnt = 0;``    ``for` `(``int` `i = 0; i < n; i++) {``        ``if` `(arr[i] == maxVal)``            ``cnt++;``    ``}` `    ``// Return the count of valid subsets``    ``return` `(``pow``(2, cnt) - 1);``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 1, 2, 1, 2 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(``int``);` `    ``cout << cntSubSets(arr, n);` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``import` `java.util.*;` `class` `GFG``{` `// Function to return the count of``// subsets with the maximum mean``static` `int` `cntSubSets(``int` `arr[], ``int` `n)``{` `    ``// Maximum value from the array``    ``int` `maxVal = Arrays.stream(arr).max().getAsInt();` `    ``// To store the number of times maximum``    ``// element appears in the array``    ``int` `cnt = ``0``;``    ``for` `(``int` `i = ``0``; i < n; i++)``    ``{``        ``if` `(arr[i] == maxVal)``            ``cnt++;``    ``}` `    ``// Return the count of valid subsets``    ``return` `(``int``) (Math.pow(``2``, cnt) - ``1``);``}` `// Driver code``public` `static` `void` `main(String []args)``{``    ``int` `arr[] = { ``1``, ``2``, ``1``, ``2` `};``    ``int` `n = arr.length;` `    ``System.out.println(cntSubSets(arr, n));``}``}` `// This code is contributed by Rajput-Ji`

## Python3

 `# Python3 implementation of the approach` `# Function to return the count of``# subsets with the maximum mean``def` `cntSubSets(arr, n) :` `    ``# Maximum value from the array``    ``maxVal ``=` `max``(arr);` `    ``# To store the number of times maximum``    ``# element appears in the array``    ``cnt ``=` `0``;``    ``for` `i ``in` `range``(n) :``        ``if` `(arr[i] ``=``=` `maxVal) :``            ``cnt ``+``=` `1``;` `    ``# Return the count of valid subsets``    ``return` `((``2` `*``*` `cnt) ``-` `1``);` `# Driver code``if` `__name__ ``=``=` `"__main__"` `:``    ` `    ``arr``=` `[ ``1``, ``2``, ``1``, ``2` `];``    ``n ``=` `len``(arr);``    ` `    ``print``(cntSubSets(arr, n));` `# This code is contributed by AnkitRai01`

## C#

 `// C# implementation of the approach``using` `System;``using` `System.Linq;``                    ` `class` `GFG``{` `// Function to return the count of``// subsets with the maximum mean``static` `int` `cntSubSets(``int` `[]arr, ``int` `n)``{` `    ``// Maximum value from the array``    ``int` `maxVal = arr.Max();` `    ``// To store the number of times maximum``    ``// element appears in the array``    ``int` `cnt = 0;``    ``for` `(``int` `i = 0; i < n; i++)``    ``{``        ``if` `(arr[i] == maxVal)``            ``cnt++;``    ``}` `    ``// Return the count of valid subsets``    ``return` `(``int``) (Math.Pow(2, cnt) - 1);``}` `// Driver code``public` `static` `void` `Main(String []args)``{``    ``int` `[]arr = { 1, 2, 1, 2 };``    ``int` `n = arr.Length;` `    ``Console.WriteLine(cntSubSets(arr, n));``}``}` `// This code is contributed by 29AjayKumar`

## Javascript

 ``
Output:
`3`

Time Complexity: O(n)

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