Número de subconjuntos cuya media es máxima

Dada una array arr[] de tamaño N , la tarea es contar el número de subconjuntos de arr[] cuya media es máxima.

Ejemplos: 

Entrada: arr[] = {1, 2, 1, 2} 
Salida: 3 
Subconjuntos con media máxima son {2}, {2} y {2, 2}.

Entrada: arr[] = {1} 
Salida: 1  

Enfoque: el valor máximo para la media de cualquier subconjunto será cuando el subconjunto solo consista en el elemento máximo de la array. Entonces, para contar todos los subconjuntos posibles, encuentre la frecuencia del elemento máximo de la array, digamos cnt , y el recuento de los subconjuntos posibles será 2 ^cnt – 1 .

A continuación se muestra la implementación del enfoque anterior:  

// 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 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 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# 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

<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>

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Espacio Auxiliar: O(1)