Sum of subsets of all the subsets of an array | O(N)
Given an array arr[] of length N, the task is to find the overall sum of subsets of all the subsets of the array.
Examples:
Input: arr[] = {1, 1}
Output: 6
All possible subsets:
a) {} : 0
All the possible subsets of this subset
will be {}, Sum = 0
b) {1} : 1
All the possible subsets of this subset
will be {} and {1}, Sum = 0 + 1 = 1
c) {1} : 1
All the possible subsets of this subset
will be {} and {1}, Sum = 0 + 1 = 1
d) {1, 1} : 4
All the possible subsets of this subset
will be {}, {1}, {1} and {1, 1}, Sum = 0 + 1 + 1 + 2 = 4
Thus, ans = 0 + 1 + 1 + 4 = 6
Input: arr[] = {1, 4, 2, 12}
Output: 513
Approach: In this article, an approach with O(N) time complexity to solve the given problem will be discussed.
The key is observing the number of times an element will repeat in all the subsets.
Let’s magnify the view. It is known that every element will appear 2(N – 1) times in the sum of subsets. Now, let’s magnify the view even further and see how the count varies with the subset size.
There are N – 1CK – 1 subsets of size K for every index that include it.
Contribution of an element for a subset of size K will be equal to 2(K – 1) times its value. Thus, total contribution of an element for all the subsets of length K will be equal to N – 1CK – 1 * 2(K – 1)
Total contribution among all the subsets will be equal to:
N – 1CN – 1 * 2(N – 1) + N – 1CN – 2 * 2(N – 2 + N – 1CN – 3 * 2(N – 3) + … + N – 1C0 * 20.
Now, the contribution of each element in the final answer is known. So, multiply it to the sum of all the elements of the array which will give the required answer.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
#define maxN 10
int fact[maxN];
int ncr( int n, int r)
{
return (fact[n] / fact[r]) / fact[n - r];
}
int findSum( int * arr, int n)
{
fact[0] = 1;
for ( int i = 1; i < n; i++)
fact[i] = i * fact[i - 1];
int mul = 0;
for ( int i = 0; i <= n - 1; i++)
mul += ( int ) pow (2, i) * ncr(n - 1, i);
int ans = 0;
for ( int i = 0; i < n; i++)
ans += mul * arr[i];
return ans;
}
int main()
{
int arr[] = { 1, 1 };
int n = sizeof (arr) / sizeof ( int );
cout << findSum(arr, n);
return 0;
}
|
Java
class GFG
{
static int maxN = 10 ;
static int []fact = new int [maxN];
static int ncr( int n, int r)
{
return (fact[n] / fact[r]) / fact[n - r];
}
static int findSum( int [] arr, int n)
{
fact[ 0 ] = 1 ;
for ( int i = 1 ; i < n; i++)
fact[i] = i * fact[i - 1 ];
int mul = 0 ;
for ( int i = 0 ; i <= n - 1 ; i++)
mul += ( int )Math.pow( 2 , i) * ncr(n - 1 , i);
int ans = 0 ;
for ( int i = 0 ; i < n; i++)
ans += mul * arr[i];
return ans;
}
public static void main(String []args)
{
int arr[] = { 1 , 1 };
int n = arr.length;
System.out.println(findSum(arr, n));
}
}
|
Python3
maxN = 10
fact = [ 0 ] * maxN;
def ncr(n, r) :
return (fact[n] / / fact[r]) / / fact[n - r];
def findSum(arr, n) :
fact[ 0 ] = 1 ;
for i in range ( 1 , n) :
fact[i] = i * fact[i - 1 ];
mul = 0 ;
for i in range (n) :
mul + = ( 2 * * i) * ncr(n - 1 , i);
ans = 0 ;
for i in range (n) :
ans + = mul * arr[i];
return ans;
if __name__ = = "__main__" :
arr = [ 1 , 1 ];
n = len (arr);
print (findSum(arr, n));
|
C#
using System;
class GFG
{
static int maxN = 10;
static int []fact = new int [maxN];
static int ncr( int n, int r)
{
return (fact[n] / fact[r]) / fact[n - r];
}
static int findSum( int [] arr, int n)
{
fact[0] = 1;
for ( int i = 1; i < n; i++)
fact[i] = i * fact[i - 1];
int mul = 0;
for ( int i = 0; i <= n - 1; i++)
mul += ( int )Math.Pow(2, i) * ncr(n - 1, i);
int ans = 0;
for ( int i = 0; i < n; i++)
ans += mul * arr[i];
return ans;
}
public static void Main(String []args)
{
int []arr = { 1, 1 };
int n = arr.Length;
Console.WriteLine(findSum(arr, n));
}
}
|
Javascript
<script>
let fact = new Array(10);
function ncr(n, r)
{
return (fact[n] / fact[r]) / fact[n - r];
}
function findSum(arr, n)
{
fact[0] = 1;
for (let i = 1; i < n; i++)
fact[i] = i * fact[i - 1];
let mul = 0;
for (let i = 0; i <= n - 1; i++)
mul += Math.pow(2, i) * ncr(n - 1, i);
let ans = 0;
for (let i = 0; i < n; i++)
ans += mul * arr[i];
return ans;
}
let arr = [ 1, 1 ];
let n = arr.length;
document.write(findSum(arr, n));
</script>
|
Time Complexity : O(Nlogn) ,where N is the number of elements in an array.
Space Complexity : O(N) ,to store the factorial of numbers from 1 to N
Last Updated :
25 Apr, 2023
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