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 A_i 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 :

DP(mask, i) = DP(mask, i-1) U DP(mask 2ⁱ, i-1)

When i-th bit is unset :

DP(mask, i) = DP(mask, i-1)

Program to illustrate the working of our solution,

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;
}