Number of subsets whose mean is maximum

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: 3 
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 2^cnt – 1.

Below is the implementation of the above approach:  

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
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

<script>

// Javascript implementation of the approach

// Function to return the count of
// subsets with the maximum mean
function cntSubSets(arr, n)
{

    // Maximum value from the array
    var maxVal =  arr.reduce(function(a, b)
    {
    return Math.max(a, b);
    });

    // To store the number of times maximum
    // element appears in the array
    var cnt = 0;
    for (var i = 0; i < n; i++) {
        if (arr[i] == maxVal)
            cnt++;
    }

    // Return the count of valid subsets
    return (Math.pow(2, cnt) - 1);
}

// Driver code
var arr = [ 1, 2, 1, 2 ]
var n = arr.length;
document.write(cntSubSets(arr, n));


</script>

Output:

3

Time Complexity: O(n)

Auxiliary Space: O(1)