# Sum over Subsets - Dynamic Programming in C++

In this problem, we are given an array arr[] of size 2n. Our task is to create a program to find the sum over subset using dynamic programming to solve it.

We need to calculate function, F(x) = Σ Ai such that x&i == i for all x. i.e. i is a bitwise subset of x.

Let’s take an example to understand the problem,

Input: A[] = {5, 7, 1, 9} , n = 2

Output: 5 12 6 22

Explanation: For n = 2 , there are 4 values of x. They are 0, 1, 2, 3.

Now, calculating values of the function:

F(0) = A0 = 5
F(1) = A0 + A1 = 5 + 7 = 12
F(2) = A0 + A2 = 5 + 1 = 6
F(3) = A0 + A1 + A2 + A3 = 5 + 7 + 1 + 9 = 22

The solution to this problem using dynamic programming, we will look at the mask and find a bitwise subset of every mask. We will store the bitwise subset using dynamic programming, which will reduce the number of repetitions. An index that has a set bit or an unset bit will be visited by 2n masks more than once.

We will recur based on the bit at i index:

When i-th bit is set :

When i-th bit is unset :

## Example

Live Demo

#include <iostream>
using namespace std;
const int N = 1000;

void SumOverSubsets(int a[], int n) {

int sum[1 << n] = {0};
int DP[N][N];
for (int i = 0; i < (1 << n); i++) {
for (int j = 0; j < n; j++) {
if (i & (1 << j)) {
if (j == 0)
DP[i][j] = a[i] + a[i ^ (1 << j)];
else
DP[i][j] = DP[i][j - 1] + DP[i ^ (1 << j)][j - 1];
} else {
if (j == 0)
DP[i][j] = a[i];
else
DP[i][j] = DP[i][j - 1];
}
}
sum[i] = DP[i][n - 1];
}
for (int i = 0; i < (1 << n); i++)
cout<<sum[i]<<"\t";
}
int main() {
int A[] = {5, 7, 1, 9};
int n = 2;
cout<<"The sum over subsets is \t";
SumOverSubsets(A, n);
return 0;
}

## Output

The sum over subsets is 5 12 6 22