Sum of the products of all possible Subsets
Last Updated :
12 Sep, 2023
Given an array of n non-negative integers. The task is to find the sum of the product of elements of all the possible subsets. It may be assumed that the numbers in subsets are small and computing product doesn’t cause arithmetic overflow.
Example :
Input : arr[] = {1, 2, 3}
Output : 23
Possible Subset are: 1, 2, 3, {1, 2}, {1, 3},
{2, 3}, {1, 2, 3}
Products of elements in above subsets :
1, 2, 3, 2, 3, 6, 6
Sum of all products = 1 + 2 + 3 + 2 + 3 + 6 + 6
= 23
Naive Approach: Simple approach is to generate all possible subset one by one and calculate sum of all elements. Time Complexity of this approach is exponential as there are total 2n – 1 subsets.
An Efficient approach is to generalize the whole problem into some pattern. Suppose we have two numbers a and b. We can write all possible subset products as:-
= a + b + ab
= a(1+b) + b + 1 - 1
= a(1+b) + (1+b) - 1
= (a + 1) * (b + 1) - 1
= (1+a) * (1 + b) - 1
Now take three numbers a, b, c:-
= a + b + c + ab + bc + ca + abc
= a + ac + b + bc + ab + abc + c + 1 - 1
= a(1+c) + b(1+c) + ab(1+c) + c + 1 - 1
= (a + b + ab + 1)(1+c) - 1
= (1+a) * (1+b) * (1+c) - 1
The above pattern can be generalized for n numbers.
Below is the implementation of above idea:
C++
#include <bits/stdc++.h>
using namespace std;
int productOfSubsetSums( int arr[], int n)
{
int ans = 1;
for ( int i = 0; i < n; ++i )
ans = ans * (arr[i] + 1);
return ans-1;
}
int main()
{
int arr[] = {1, 2, 3, 4};
int n = sizeof (arr)/ sizeof arr[0];
cout << productOfSubsetSums(arr, n);
return 0;
}
|
Java
public class Subset
{
static int productOfSubsetSums( int arr[], int n)
{
int ans = 1 ;
for ( int i = 0 ; i < n; ++i )
ans = ans * (arr[i] + 1 );
return ans- 1 ;
}
public static void main (String[] args)
{
int arr[] = { 1 , 2 , 3 , 4 };
int n = arr.length;
System.out.println(productOfSubsetSums(arr, n));
}
}
|
Python3
def productOfSubsetSums(arr, n):
ans = 1 ;
for i in range ( 0 ,n):
ans = ans * (arr[i] + 1 )
return ans - 1
arr = [ 1 , 2 , 3 , 4 ]
n = len (arr)
print (productOfSubsetSums(arr, n))
|
C#
using System;
public class Subset
{
static int productOfSubsetSums( int []arr, int n)
{
int ans = 1;
for ( int i = 0; i < n; ++i )
ans = ans * (arr[i] + 1);
return ans-1;
}
public static void Main ()
{
int []arr = {1, 2, 3, 4};
int n = arr.Length;
Console.Write(productOfSubsetSums(arr, n));
}
}
|
PHP
<?php
function productOfSubsetSums( $arr , $n )
{
$ans = 1;
for ( $i = 0; $i < $n ; ++ $i )
$ans = $ans * ( $arr [ $i ] + 1);
return $ans -1;
}
$arr = array (1, 2, 3, 4);
$n = sizeof( $arr );
echo (productOfSubsetSums( $arr , $n ));
?>
|
Javascript
<script>
function productOfSubsetSums(arr, n)
{
let ans = 1;
for (let i = 0; i < n; ++i )
ans = ans * (arr[i] + 1);
return ans-1;
}
let arr = [1, 2, 3, 4];
let n = arr.length;
document.write(productOfSubsetSums(arr, n));
</script>
|
Time Complexity: O(n)
Auxiliary Space: O(1)
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