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.
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.
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))
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 subsetsint 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 methodsint 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 arrayclass 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 subsetsdef 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 codearr = [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 arrayusing 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 |
PHP
<?php// PHP code to find sum of maximum of// all subsets of array// Method returns sum of maximum of all subsetsfunction 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?> |
Javascript
<script>// JavaScript code to find sum of// maximum of all subsets of array// Method returns sum of maximum of all subsetsfunction 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> |
Output:
28
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