Sum of maximum elements of all subsets

Given an array of integer numbers, we need to find sum of maximum number of all possible subsets.

Examples:

Input  : arr = {3, 2, 5}
Output : 28
Explanation : 
Subsets and their maximum are,
{}            maximum = 0
{3}             maximum = 3
{2}            maximum = 2
{5}            maximum = 5
{3, 2}            maximum = 3
{3, 5}            maximum = 5
{2, 5}            maximum = 5
{3, 2, 5}        maximum = 5
Sum of maximum will be, 0 + 3 + 2 + 5 + 3 + 5 + 5 + 5 = 28,
which will be our answer.

 Brute force method:

A simple solution is to iterate through all subsets of array and finding maximum of all of them and then adding them in our answer, but this approach will lead us to exponential time complexity.

Approach:

We first define an array arr containing the input integers and find its size n. We also initialize a variable sum to 0, which we will use to store the sum of maximums of all subsets.

Next, we use a nested loop. The outer loop runs from 0 to (1 << n) – 1, where (1 << n) is the number of possible subsets of an array of size n. The inner loop runs from 0 to n-1, checking if the jth element of the array is present in the current subset or not. This is done by checking if the jth bit of the binary representation of i is set or not. If it is set, we update the maximum number of the subset.

After finding the maximum number of each subset, we add it to the sum variable. Finally, we print the sum as the output.

Implementation:

C++

#include <bits/stdc++.h>
using namespace std;
int sumOfMaximumOfSubsets(int arr[],int n)
{
    int sum=0;
    for(int i = 0; i < (1 << n); i++) {
        int max_num = 0;
        // loop through all elements of the subset
        for(int j = 0; j < n; j++) {
            if(i & (1 << j)) {
                max_num = max(max_num, arr[j]);
            }
        }
        sum += max_num;
    }
    return sum;
}

int main() {
    int arr[] = {3,2, 5};
    int n = sizeof(arr)/sizeof(arr[0]);
    cout << sumOfMaximumOfSubsets(arr, n) << endl;
    return 0;
}

Java

public class GFG {
    // Function to calculate the sum of maximum elements of all subsets
    static int sumOfMaximumOfSubsets(int[] arr, int n) {
        int sum = 0;
        // Iterate through all possible subsets using bit manipulation
        for (int i = 0; i < (1 << n); i++) {
            int max_num = 0;
            // Loop through all elements of the subset
            for (int j = 0; j < n; j++) {
                if ((i & (1 << j)) != 0) {
                    max_num = Math.max(max_num, arr[j]);
                }
            }
            sum += max_num;
        }
        return sum;
    }

    public static void main(String[] args) {
        int[] arr = { 3, 2, 5 };
        int n = arr.length;
        System.out.println(sumOfMaximumOfSubsets(arr, n));
    }
}

Python

def sum_of_maximum_of_subsets(arr):
    n = len(arr)
    total_sum = 0

    # Loop through all possible subsets
    for i in range(1 << n):
        max_num = 0

        # Loop through all elements of the subset
        for j in range(n):
            if i & (1 << j):
                max_num = max(max_num, arr[j])

        total_sum += max_num

    return total_sum

# Main function


def main():
    arr = [3, 2, 5]
    result = sum_of_maximum_of_subsets(arr)
    print(result)


if __name__ == "__main__":
    main()

C#

using System;
public class GFG
{
    public static int SumOfMaximumOfSubsets(int[] arr, int n)
    {
        int sum = 0;
        for (int i = 0; i < (1 << n); i++)
        {
            int maxNum = 0;
            // Loop through all elements of the subset
            for (int j = 0; j < n; j++)
            {
                if ((i & (1 << j)) != 0)
                {
                    maxNum = Math.Max(maxNum, arr[j]);
                }
            }
            sum += maxNum;
        }
        return sum;
    }
    public static void Main()
    {
        int[] arr = { 3, 2, 5 };
        int n = arr.Length;
        Console.WriteLine(SumOfMaximumOfSubsets(arr, n));
    }
}

Javascript

function sumOfMaximumOfSubsets(arr, n)
{
    let sum=0;
    for(let i = 0; i < (1 << n); i++) {
        let max_num = 0;
        // loop through all elements of the subset
        for(let j = 0; j < n; j++) {
            if(i & (1 << j)) {
                max_num = Math.max(max_num, arr[j]);
            }
        }
        sum += max_num;
    }
    return sum;
}

let arr = [3,2, 5];
let n = arr.length;
console.log(sumOfMaximumOfSubsets(arr, n));

output:

28

Time Complexity: O(2^n * n), where n is the size of the input array.
Auxiliary Space: O(1), since no extra space being used.

An efficient solution is based on one thing, how many subsets of array have a particular element as their maximum. As in above example, four subsets have 5 as their maximum, two subsets have 3 as their maximum and one subset has 2 as its maximum. The idea is to compute these frequencies corresponding to each element of array. Once we have frequencies, we can just multiply them with array values and sum them all, which will lead to our final result. 

To find frequencies, first we sort the array in non-increasing order and when we are standing at a[i] we know, all element from a[i + 1] to a[N-1] are smaller than a[i], so any subset made by these element will choose a[i] as its maximum so count of such subsets corresponding to a[i] will be, 2^(N – i – 1) (total subset made by array elements from a[i + 1] to a[N]). 

If same procedure is applied for all elements of array, we will get our final answer as, 
res = a[0]*2^(N-1) + a[1]*2^(N-2) ….. + a[i]*2^(N-i-1) + ….. + a[N-1]*2^(0) 
Now if we solve above equation as it is, calculating powers of 2 will take time at each index, 
instead we can reform the equation similar to horner’s rule for simplification, 
res = a[N] + 2*(a[N-1] + 2*(a[N-2] + 2*( …… 2*(a[2] + 2*a[1])…..)))) 
Total complexity of above solution will be O(N*log(N)) 

Implementation:

C++

// C/C++ code to find sum of maximum of all subsets of array
#include <bits/stdc++.h>
using namespace std;

// Method returns sum of maximum of all subsets
int sumOfMaximumOfSubsets(int arr[], int N)
{
    //    sorting array in decreasing order
    sort(arr, arr + N, greater<int>());

    //    initializing sum with first element
    int sum = arr[0];
    for (int i = 1; i < N; i++)
    {
        // calculating evaluation similar to horner's rule
        sum = 2 * sum + arr[i];
    }

    return sum;
}

// Driver code to test above methods
int main()
{
    int arr[] = {3, 2, 5};
    int N = sizeof(arr) / sizeof(arr[0]);
    cout << sumOfMaximumOfSubsets(arr, N) << endl;
    return 0;
}

Java

import java.util.Arrays;
import java.util.Collections;

// Java code to find sum of
// maximum of all subsets of array
class GFG
{

    // Method returns sum of maximum of all subsets
    static int sumOfMaximumOfSubsets(Integer arr[], int N)
    {
        // sorting array in decreasing order
        Arrays.sort(arr, Collections.reverseOrder());

        // initializing sum with first element
        int sum = arr[0];
        for (int i = 1; i < N; i++)
        {
            // calculating evaluation similar to horner's rule
            sum = 2 * sum + arr[i];
        }

        return sum;
    }

    // Driver code
    public static void main(String[] args)
    {
        Integer arr[] = {3, 2, 5};
        int N = arr.length;
        System.out.println(sumOfMaximumOfSubsets(arr, N));
    }
}

/* This code contributed by PrinciRaj1992 */

Python3

# Python 3 code to find sum
# of maximum of all subsets
# of array

# Method returns sum of
# maximum of all subsets
def sumOfMaximumOfSubsets(arr, N):

    # sorting array in
    # decreasing order
    arr.sort(reverse = True)

    # initializing sum
    # with first element
    sum = arr[0]
    for i in range(1, N):

        # calculating evaluation
        # similar to horner's rule
        sum = 2 * sum + arr[i]

    return sum

# Driver code
arr = [3, 2, 5]
N = len(arr)
print(sumOfMaximumOfSubsets(arr, N))

# This code is contributed
# by Smitha

C#

// C# code to find sum of
// maximum of all subsets of array
using System;

class GFG
{

    // Method returns sum of maximum of all subsets
    static int sumOfMaximumOfSubsets(int []arr, int N)
    {
        // sorting array in decreasing order
        Array.Sort(arr);
        Array.Reverse(arr);

        // initializing sum with first element
        int sum = arr[0];
        for (int i = 1; i < N; i++)
        {
            // calculating evaluation
            // similar to horner's rule
            sum = 2 * sum + arr[i];
        }

        return sum;
    }

    // Driver code
    public static void Main(String[] args)
    {
        int []arr = {3, 2, 5};
        int N = arr.Length;
        Console.WriteLine(sumOfMaximumOfSubsets(arr, N));
    }
}

// This code has been contributed by 29AjayKumar

Javascript

<script>

// JavaScript code to find sum of
// maximum of all subsets of array

// Method returns sum of maximum of all subsets
function sumOfMaximumOfSubsets(arr,N)
{
    // sorting array in decreasing order

    arr.sort((a,b)=>a-b);
    arr.reverse();
    // initializing sum with first element
    let sum = arr[0];
    for (let i = 1; i < N; i++)
    {
        // calculating evaluation similar to horner's rule
        sum = 2 * sum + arr[i];
    }

    return sum;
}


    let arr= [3, 2, 5];
    let N = arr.length;
    document.write(sumOfMaximumOfSubsets(arr, N));

</script>

PHP

<?php
// PHP code to find sum of maximum of
// all subsets of array

// Method returns sum of maximum of all subsets
function sumOfMaximumOfSubsets($arr, $N)
{
    // sorting array in decreasing order
    rsort($arr);

    // initializing sum with first element
    $sum = $arr[0];
    for ( $i = 1; $i < $N; $i++)
    {
        // calculating evaluation similar
        // to horner's rule
        $sum = 2 * $sum + $arr[$i];
    }

    return $sum;
}

// Driver Code
$arr = array(3, 2, 5);
$N = count($arr);
echo sumOfMaximumOfSubsets($arr, $N);

// This code is contributed by Rajput-Ji
?>

Output

28

Time Complexity: O(N log N), since sort() function is used.
Auxiliary Space: O(1), since no extra space being used.